Latihan Soal 8 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 66.&\textrm{Suatu unit pekerjaan dapat diselesaikan oleh A}\\ &\textrm{B, dan C bersama-sama dalam 2 jam saja.}\\ &\textrm{Jika pekerjaan itu dapat diselesaikan oleh A dan}\\ &\textrm{B bersama-sama dalam 2 jam 24 menit, dan oleh}\\ &\textrm{B dan C bersama-sama dalam waktu 3 jam,}\\ &\textrm{maka sistem persamaan berikut yang memenuhi}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\begin{cases} A+B+C&=2 \\ A+B & =\displaystyle \frac{12}{5} \\ B+C &=3 \end{cases}\\ \textrm{b}.&\begin{cases} A+B+C&=\displaystyle \frac{1}{2} \\ A+B & =\displaystyle \frac{5}{12} \\ B+C &=\displaystyle \frac{1}{3} \end{cases}\\ \textrm{c}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=2 \\ \displaystyle \frac{1}{A}+\frac{1}{B}& =\displaystyle \frac{12}{5} \\ \displaystyle \frac{1}{B}+\frac{1}{C}&=3 \end{cases}\\ \color{red}\textrm{d}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=\displaystyle \frac{1}{2} \\ \displaystyle \frac{1}{A}+\frac{1}{B}& =\displaystyle \frac{5}{12} \\ \displaystyle \frac{1}{B}+\frac{1}{C}&=\displaystyle \frac{1}{3} \end{cases}\\ \textrm{e}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=2 \\ \displaystyle \frac{1}{A}+\frac{1}{B}-\frac{1}{C}& =\displaystyle \frac{12}{5} \\ \displaystyle -\frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=3 \end{cases} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Per}&\textrm{hatikan bahwa}:\color{red}\textrm{Waktu penyelesaian}\\ \color{red}\textrm{sua}&\color{red}\textrm{tu pekerjaan adalah termasuk}\\ \color{red}\textrm{per}&\color{red}\textrm{bandingan berbalik nilai},\: \color{blue}\textrm{maka}\\ \bullet \: \: \: &A,B,\: \textrm{dan}\: C \: \textrm{dalam 2 jam, artinya}:\\ &\color{black}\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{2},\: \color{blue}\textrm{demikian juga}\\ \bullet \: \: \: &A\: \textrm{dan}\: B\: \textrm{bersama-sama selesai dalam}\\ &\textrm{2 jam 24 menit atau}\: \displaystyle \frac{12}{5}\: \textrm{jam}:\\ &\color{black}\displaystyle \frac{1}{A}+\frac{1}{B}=\frac{5}{12}\\ \bullet \: \: \: &B\: \textrm{dan}\: C\: \textrm{selesai dalam 3 jam}:\\ &\color{black}\displaystyle \frac{1}{B}+\frac{1}{C}=\frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 67.&\textrm{Himpunan penyelesaian dari}\\ &\left\{\begin{matrix} x+y+4z=15\quad\\ x-y+z=2\qquad\\ x+2y-3z=-4 \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ (-1,1,3) \right \}\\ \color{red}\textrm{b}.&\left \{ (1,2,3) \right \}\\ \textrm{c}.&\left \{ (-2,1,1) \right \}\\ \textrm{d}.&\left \{ (3,2,-1) \right \}\\ \textrm{e}.&\left \{ (1,-2,3) \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Semunya dikerjakan dengan metode}\\ &\color{blue}\textrm{matriks}\: (\color{black}\textbf{Cara Cramer})\\ &\begin{aligned} \color{blue}x&=\displaystyle \frac{\begin{vmatrix} 15 & 1 & 4\\ 2& -1 & 1\\ -4& 2 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{15\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}+4\begin{vmatrix} 2 & -1\\ -4 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{15(3-2)-1(-6+4)+4(4-4)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{15(1)-1(-2)+4(0)}{1(1)-1(-4)+4(3)}=\frac{17}{17}=1 \\ \color{blue}y&=\displaystyle \frac{\begin{vmatrix} 1 & 15 & 4\\ 1& 2 & 1\\ 1& -4 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}-15\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(-6+4)-15(-3-1)+4(-4-2)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(-2)-15(-4)+4(-6)}{1(1)-1(-4)+4(3)}=\frac{34}{17}=2\\ \color{blue}z&=\displaystyle \frac{\begin{vmatrix} 1 & 1 & 15\\ 1& -1 & 2\\ 1& 2 & -4 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} -1 & 2\\ 2 & -4 \end{vmatrix}-1\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}+15\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(4-4)-1(-4-2)+15(2+1)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(0)-1(-6)+15(3)}{1(1)-1(-4)+4(3)}=\frac{51}{17}=3 \end{aligned} \end{array}$

$.\quad\quad \color{blue}\textrm{Cara di atas}$  full matriks-Cramer

$\begin{array}{ll}\\ 68.&\textrm{Hasil dari}\: \: xyz\: \: \textrm{yang memenuhi}\\ &\left\{\begin{matrix} x+y+z=2\quad\\ x-y+z=-2\: \\ x-y-z=2\quad \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-8\\ \textrm{b}.&-4\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=2\quad.....(1)\\ x-y+z=-2\: .....(2)\\ x-y-z=2\quad .....(3) \end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llc}\\ x+y+z&=2&\\ x-y+z&=-2&-\\\hline \: \, \quad2y&=4&\\ \qquad\quad y&=2&....(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lcc}\\ x+y+z&=2&\\ x-y-z&=2&+\\\hline 2x&=4&\\ \qquad\quad x&=2&....(5) \end{array} \\ &\textrm{Persamaan}\: \: (4)\&(5)\: \: \textrm{ke}\: \: (1)\\ &\color{red}\begin{aligned}x+y+z&=2\\ (2)+(2)+z&=2\\ z&=-2 \end{aligned}\\ &\textrm{Jadi},\: \: xyz=(2).(2).(-2)=\color{red}-8 \end{aligned} \end{array}$

$.\quad\: \:  \color{black}\textrm{Cara di atas}$  full eliminasi-substitusi

$\begin{array}{ll}\\ 69.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y+z=-6\quad\\ x-2y+z=3\quad\: \\ -2x+y+z=9\quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: xyz=....\\ &\begin{array}{llll}\\ \textrm{a}.&-30\\ \textrm{b}.&-15\\ \textrm{c}.&5\\ \color{red}\textrm{d}.&30\\ \textrm{e}.&35 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=-6\quad ....(1)\\ x-2y+z=3\quad\: ....(2)\\ -2x+y+z=9\quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llc}\\ x+y+z&=-6&\\ x-2y+z&=3&-\\\hline \: \: \: \quad 3y&=-9&\\ \qquad\quad y&=-3&....(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llc}\\ x+y+z&=-6&\\ -2x+y+z&=9&-\\\hline 3x&=-15&\\ \qquad\quad x&=-5&....(5) \end{array} \\ &\textrm{Persamaan}\: \: (4)\&(5)\: \: \textrm{ke}\: \: (1)\\ &\color{red}\begin{aligned}x+y+z&=2\\ (-5)+(-3)+z&=-6\\ z&=2 \end{aligned}\\ &\textrm{Jadi},\: \: xyz=(-5).(-3).(2)=\color{red}30 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 70.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+2y+z=4\: \: \qquad\\ 3x+y+2z=-5\quad\: \\ x-2y+2z=-6\quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: xyz=....\\ &\begin{array}{llll}\\ \textrm{a}.&-96\\ \color{red}\textrm{b}.&-24\\ \textrm{c}.&24\\ \textrm{d}.&32\\ \textrm{e}.&96 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+2y+z=4\: \qquad.......(1)\\ 3x+y+2z=-5\quad\: ......(2)\\ x-2y+2z=-6\quad .......(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llclll}\\ x+2y+z&=4&\left | \times 1 \right |&\: \: x+2y+z&=4\\ 3x+y+2z&=-5&\left | \times 2 \right |&6x+2y+4z&=-10&-\\\hline &&&-5x\: \: \quad-3z&=14&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lll}\\ x+2y+z&=4&\\ x-2y+2z&=-6&+\\\hline 2x\: \: \: \, \quad +3z&=-2&...(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\begin{array}{lll}\\ -5x-3z&=14&\\ 2x+3z&=-2&+\\\hline -3x&=12&\\ \qquad\quad x&=-4&.....(6)\\ \color{red}\textrm{didapat pula}&z&=2......(7) \end{array}\\ &\textrm{Dari persamaan}\: \: (6)\&(7)\: \: \textrm{didapatkan}\\ &\color{red}\begin{aligned}x+2y+z&=4\\ (-4)+2y+2&=4\\ y&=3 \end{aligned}\\ &\textrm{Jadi},\: \: xyz=(-4).(3).(2)=\color{red}-24 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 71.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y-z=1\: \: \qquad\\ 2x-y+2z=9\quad\: \\ x+3y-z=7\: \: \: \: \quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: \displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{3}{4}\\ \color{red}\textrm{c}.&\displaystyle \frac{13}{12}\\ \textrm{d}.&\displaystyle \frac{5}{4}\\ \textrm{e}.&\displaystyle \frac{7}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y-z=1\: \: \qquad....(1)\\ 2x-y+2z=9\quad\: ....(2)\\ x+3y-z=7\: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ x+y-z&=1&\\ 2x-y+2z&=9&+\\\hline 3x\: \: \qquad+z&=10&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ x+y-z&=1&\left | \times 3 \right |&3x+3y-3z&=3&\\ x+3y-z&=7&\left | \times 1 \right |&\quad x+3y-z&=7&-\\\hline &&&2x\quad \: \: \quad-2z&=-4&\\ &&&\: \: x\quad \: \: \: \: \quad-z&=2&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\begin{array}{lll}\\ 3x+z&=10&\\ x-z&=-2&+\\\hline 4x&=8&\\ \qquad\quad x&=2&.....(6)\\ \color{red}\textrm{didapat pula}&z&=4......(7) \end{array} \\ &\textrm{Dari persamaan}\: \: (1)\&(3)\: \: \textrm{didapatkan juga}\\ &\begin{array}{lll}\\ x+y-z&=1&\\ x+3y-z&=-7&-\\\hline \quad -2y&=-6&\\ \qquad\qquad y&=3&....(8) \end{array}\\ &\textrm{Jadi},\: \: \displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\color{red}\frac{13}{12} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 72.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y+z=5\: \: \: \: \quad\\ x+y-4z=10\quad \\ -2x+y+z=0 \quad \end{matrix}\right.\\ &\textrm{Nilai dari}\: \: \displaystyle \frac{xz}{y}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\frac{6}{13}\\ \color{red}\textrm{b}.&\displaystyle -\frac{5}{13}\\ \textrm{c}.&\displaystyle -\frac{1}{13}\\ \textrm{d}.&\displaystyle \frac{1}{13}\\ \textrm{e}.&\displaystyle \frac{7}{13} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=5\: \: \: \: \quad.....(1)\\ x+y-4z=10\quad .....(2)\\ -2x+y+z=0 \quad .....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ x+y+z&=5&\\ x+y-4z&=10&-\\\hline \: \: \qquad \: \: \: \: \: 5z&=-5&\\ \: \: \qquad\quad \: \: \: z&=-1&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ x+y+z&=5&\\ -2x+y+z&=0&-\\\hline 3x\quad \: \quad&=5&\\ \: \: \quad \: \: \: \: \quad x&=\displaystyle \frac{5}{3}&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{aligned}x+y+z&=5\\ \displaystyle \frac{5}{3}+y-1&=5\\ y&=5+1-\displaystyle \frac{5}{3}=\frac{13}{3} \end{aligned}\\ &\textrm{Jadi},\: \: \displaystyle \frac{xz}{y}=\displaystyle \frac{\left ( \displaystyle \frac{5}{3} \right ).(-1)}{\displaystyle \frac{13}{3}}=\color{red}-\displaystyle \frac{5}{13} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 73.&\textrm{Himpunan penyelesaian dari}\\ &\begin{cases} \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z} & =8 \\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z} & =10 \\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z} & =4 \end{cases}\\ &\textrm{adalah}\: \: \left \{ (x,y,z) \right \},\: \textrm{maka}\: \: x+3z=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&1\\ \color{red}\textrm{d}.&3\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\begin{cases} \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z} & =8\: ....(1)\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z} & =10\: .....(2)\\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z} & =4\: ...........(3) \end{cases}\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}&=8\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}&=10&-\\\hline -\displaystyle \frac{1}{x}\: \: \: \: \: -\frac{1}{z}&=-2\\ \displaystyle \frac{1}{x}\: \: \: \: \: \: \: \: +\frac{1}{z}&=2&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lllllll}\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}&=8&\left | \times 2 \right |&\displaystyle \frac{4}{x}+\frac{4}{y}+\frac{8}{z}&=16\\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}&=4&\left | \times 1 \right |&\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}&=4&-\\\hline &&&\displaystyle \frac{2}{x}\: \: \: \: \: +\frac{6}{z}&=12\\ &&\Leftrightarrow &\displaystyle \frac{1}{x}\: \: \: \: \: +\frac{3}{z}&=6&...(5)\\ \end{array}\\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\begin{array}{llll}\\ \displaystyle \frac{1}{x}+\frac{3}{z}&=6\\ \displaystyle \frac{1}{x}+\frac{1}{z}&=2\: \: \: -\\\hline \qquad\displaystyle \frac{2}{z}&=4&\\ \qquad z&=\displaystyle \frac{1}{2}\: \: ......(6)\\ \qquad x&=2-\displaystyle \frac{1}{2}=\displaystyle \frac{3}{2} \end{array}\\ &\textrm{Jadi},\: \: x+3z=\displaystyle \frac{3}{2}+3.\frac{1}{2}=\color{red}3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 74.&\textrm{Diketahui tiga buah bilangan berturut-turut}\\ &a,\: b,\: \textrm{dan}\: c.\: \textrm{Rata-rata dari ke tiga bilangan}\\ &\textrm{itu adalah 12. Bilangan kedua sama dengan}\\ &\textrm{jumlah bilangan yang lain dikurangi 12}.\\ &\textrm{Jika bilangan ke tiga sama dengan jumlah}\\ &\textrm{bilangan yang lain, maka nilai}\: \: 2a+b-c=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle 42\\ \textrm{b}.&-\displaystyle 36\\ \textrm{c}.&-\displaystyle 18\\ \textrm{d}.&-\displaystyle 12\\ \color{red}\textrm{e}.&-\displaystyle 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Model matematika dari persamaan di atas}\\ &\left\{\begin{matrix} a+b+c=36\: \: \qquad....(1)\\ -a+b-x=12\quad\: ....(2)\\ a+b-c=0\: \: \: \: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ a+b+c&=36&\\ -a+b-c&=12&+\\\hline \: \: \: \: \: \: \: \: \: \: \: 2b&=48&\\ \: \: \qquad\quad \: \: \: b&=24&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ a+b+c&=36&\\ a+b-c&=0&-\\\hline \quad\qquad2c &=36&\\ \: \: \quad \: \: \: \: \quad c&=18&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{aligned}a+b+c&=36\\ a+24+18&=36\\ a&=36-42\\ &=-6 \end{aligned} \\ &\textrm{Jadi},\: \: 2a+b-c=2(-6)+24-18=\color{red}-6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 75.&\textrm{Jumlah uang terdiri atas koin pecahan}\: \: Rp500,00\\ &Rp200,00\: \: dan\: \: Rp100,00\: \: \textrm{dengan nilai total}\\ &Rp100.000,00.\: \textrm{Jika nilai uang pecahan 500-an}\\ &\textrm{setengah dari nilai uang pecahan 200-an, tetapi}\\ &\textrm{tiga kali uang pecahan 100-an, maka banyak koin}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&460\\ \textrm{b}.&440\\ \textrm{c}.&420\\ \textrm{d}.&380\\ \textrm{e}.&350 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Model matematika dari kasus di atas}\\ &\left\{\begin{matrix} A(500)+B(200)+C(100)=100.000\: ....(1)\\ A(500)=\displaystyle \frac{1}{2}B(200)\qquad\qquad\qquad\qquad\: ....(2)\\ A(500)=3C(100)\qquad\qquad\qquad\: \: \, \: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Dari persamaan}\: \: (2)\: \textrm{didapatkan}\\ &2A(500)=B(200)\\ &\textrm{Dari persamaan}\: \: (3)\: \textrm{akan didapatkan}\\ &\displaystyle \frac{1}{3}A(500)=C(100)\\ &\textrm{Dari persamaan}\: \: (1)\: \: \textrm{maka},\\ &A(500)+B(200)+C(100)=100.000\\ &A(500)+2A(500)+\displaystyle \frac{1}{3}A(500)=100.000\\ &\displaystyle \frac{10}{3}A(500)=100.000\Leftrightarrow A(500)=30.000\\ &\textrm{maka akan didapatkan}\\ &B(200)=2(30.000)=60.000\\ &C(100)=\displaystyle \frac{1}{3}(30.000)=10.000\\ &\color{red}\begin{cases} A(500) &=30.000\Rightarrow \color{black}A=\displaystyle \frac{30.000}{500}=60 \\ B(200) &=60.000\Rightarrow \color{black}B=\displaystyle \frac{60.000}{200}=300 \\ C(100) &=10.000\Rightarrow \color{black}C=\displaystyle \frac{10.000}{100}=100 \end{cases}\\ &\textrm{Jadi},\: \: A+B+C=60+300+100=\color{red}460 \end{aligned} \end{array}$.

Latihan Soal 7 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{l}\\ 56.&\textrm{Nilai}\: \: x\: \: \textrm{berikut yang tidak memenuhi}\\ &\displaystyle \frac{x-3}{x^{2}+2x+1}\leq 0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \color{red}\textrm{b}.&-1\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\displaystyle \frac{x-3}{x^{2}+2x+1}\leq 0\\ &\displaystyle \frac{(x-3)}{(x+1)^{2}}\leq 0\\ &\textrm{Pembuat nol}\\ &\begin{cases} x =3 ,\: \textrm{boleh digunakan}\\ x =-1,\: \textrm{tetapi}\: \: x\neq -1, \end{cases}\\ &\textrm{sehingga}\: \: -1\: \: \textrm{tidak digunakan}\\ &\color{red}\begin{array}{ccc|cccc|ccccc}\\ &&&&&&&&&&\\ &-&-&-&-&-&&+&+&&\\\hline &&-1&&&&3&&&&\\ &&&&&&&&&&\\ \end{array}& \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 57.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{6x-5}\leq x \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1<x<0\\\\ \textrm{b}.&x<1\: \: \textrm{atau}\: \: x\geq 5\\\\ \color{red}\textrm{c}.&\displaystyle \frac{5}{6}\leq x\leq 1\: \: \textrm{atau}\: \: x\geq 5\\\\ \textrm{d}.&\displaystyle \frac{5}{6}\leq x< 1\: \: \textrm{atau}\: \: 5<x<6\\\\ \textrm{e}.&x\geq 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\sqrt{6x-5}&\leq x\\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 6x-5&\leq x^{2}\\ -x^{2}+6x-5&\leq 0\\ x^{2}-6x+5&\geq 0\\ (x-1)(x-5)&\geq 0\\ x\leq 1\: \: \textrm{atau}&\: \: x\geq 5\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 6x-5&\geq 0\\ 6x&\geq 5\\ x&\geq \displaystyle \frac{5}{6} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 58.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{6x+6}>6 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x>7\\ \textrm{b}.&x\geq 7\\ \textrm{c}.&x<7\\ \textrm{d}.&x>1\\ \textrm{e}.&x\geq 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\sqrt{6x+6}&>6\\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 6x+6&>36\\ x+1&>6\\ x&>7\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 6x+6&\geq 0\\ 6x&\geq 6\\ x&\geq \displaystyle \frac{6}{6}\\ x&\geq 1 \end{aligned} \end{array}$

$\begin{array}{l}\\ 59.&\textrm{Penyelesaian pertidaksamaan}\\ &x+2>\displaystyle \sqrt{10-x^{2}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\leq x\leq \sqrt{10}\\ \color{red}\textrm{b}.&1<x\leq \sqrt{10}\\ \textrm{c}.&-3<x\leq \sqrt{10}\\ \textrm{d}.&-\sqrt{10}\leq x\leq \sqrt{10}\\ \textrm{e}.&x< -3\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}x+2&>\displaystyle \sqrt{10-x^{2}} \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ x^{2}+4x+4&>10-x^{2}\\ 2x^{2}+4x&+4-10>0\\ 2x^{2}+4x-&6>0\\ x^{2}+2x-&3>0\\ (x+3)&(x-1)>0\\ x<-3&\: \: \textrm{atau}\: \: x>1\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 10-x^{2}&\geq 0\\ x^{2}-10&\leq 0\\ (x-\sqrt{10})&(x+\sqrt{10})\leq 0\\ -\sqrt{10}\leq x&\leq \sqrt{10} \end{aligned} \end{array}$

$\begin{array}{l}\\ 60.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{3x+7}\geq x-1 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<6\\ \textrm{b}.&-1\leq x<6\\ \textrm{c}.&x\geq -\displaystyle \frac{7}{3}\\ \color{red}\textrm{d}.&-\displaystyle \frac{7}{3}\leq x\leq 6\\ \textrm{e}.&-\displaystyle \frac{7}{3}\leq x\leq 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\sqrt{3x+7}&\geq x-1 \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 3x+7&\geq x^{2}-2x+1\\ -x^{2}+3x&+2x+7-1\geq 0\\ -x^{2}+5x+&6\geq 0\\ x^{2}-5x-&6\leq 0\\ (x+1)&(x-6)\leq 0\\ -1\leq x&\leq 6\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 3x+7&\geq 0\\ 3x&\geq -7\\ x&\geq -\displaystyle \frac{7}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 61.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & \sqrt{2x-8}<\sqrt{x+5}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x\geq -5\\ \textrm{b}.&x<-13\: \: \textrm{atau}\: \: x\geq -4\\ \textrm{c}.&x<13\\ \color{red}\textrm{d}.&4\leq x< 13\\ \textrm{e}.&-5\leq x\leq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\sqrt{2x-8}&<\sqrt{x+5}\\ (1)\: \: \, \textrm{kuadratkan}&\\ 2x-8&<x+5\\ x&<13\\ (2)\quad 2x-8\geq 0&\\ x&\geq 4\\ (3)\: \: \quad x+5\geq 0&\\ x&\geq -5\\ \textrm{perhatikan}&\textrm{lah garis bilangannya berikut}\\  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 62.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{6x-4}< \sqrt{2x+8} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-4<x\leq \displaystyle \frac{2}{3}\\ \textrm{b}.&-4<x<3\\ \color{red}\textrm{c}.&\displaystyle \frac{2}{3}\leq x< 3\\ \textrm{d}.&2<x\leq 4\\ \textrm{e}.&-4\leq x\leq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\displaystyle \sqrt{6x-4}&< \sqrt{2x+8} \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 6x-4&<2x+8\\ 6x-2x&<8+4\\ 4x&<12\\ x&<3\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 6x-4&\geq 0\\ 6x&\geq 4\\ x&\geq \displaystyle \frac{2}{3}\\ 3.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 2x+8&\geq 0\\ 2x&\geq -8\\ x&\geq -4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 63.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{x+3}> \sqrt{12-2x} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3<x\leq 6\\ \color{red}\textrm{b}.&-3<x\leq 6\\ \textrm{c}.&-6<x\leq 3\\ \textrm{d}.&-6<x\leq -3\\ \textrm{e}.&x<3\: \: \textrm{atau}\: \: x>6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \displaystyle \sqrt{x+3}&> \sqrt{12-2x} \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ x+3&>12-2x\\ x+2x&>12-3\\ 3x&>9\\ x&>3\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ x+3&\geq 0\\ x&\geq -3\\ 3.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 12-2x&\geq 0\\ 2x-12&\leq 0\\ 2x&\leq 12\\ x&\leq 6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 64.&(\textbf{SBMPTN 2013 Mat Das})\\ &\textrm{Jika}\: \: 1<m<2\: ,\: \textrm{maka semua nilai}\\ &x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}+4x}{-x^{2}+3x-3m}>0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x>-3\\ \textrm{b}.&x<-4\\ \color{red}\textrm{c}.&-4<x<0\\ \textrm{d}.&x<-4\: \: \textrm{atau}\: \: x>0\\ \textrm{e}.&x<-3\: \: \textrm{atau}\: \: x>-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}1.\quad&\textrm{Diketahui bahwa}:\: \: \color{red}\displaystyle \frac{x^{2}+4x}{-x^{2}+3x-3m}>0\\ &\color{black}\textrm{dengan kondisi}\: \: 1<m<2\\ &\textrm{Perhatikanlah penyebutnya yang}\\ &\textrm{mengandung bilangan}\: \: \color{red}m\: \: \color{blue}\textrm{yang terletak}\\ & \textrm{pada interval}\: :\: \color{red}1<m<2.\\ 2.\quad&\textrm{Kita cek kondisi penyebutnya dengan}\\ &\textrm{menentukan}\: Diskriminan(D)-\textrm{nya}\\ &\textrm{yaitu}:\\ &\color{red}ax^{2}+bx+c \color{black}\begin{cases} \color{black}a>0\: \&\: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit positif} \\ \color{red}a<0\: \& \: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit negatif} \end{cases}\\ &\textrm{Karena penyebutnya}\: :\: -x^{2}+3x-3m,\\ &\textrm{dengan}\: \color{red}a=-1,\: b=3,\: \&\: c=-3m,\: \color{blue}\textrm{maka}\\ &\color{black}D=\color{red}3^{2}-4(-1)(-3m)=\color{black}9-12m\\ 3.\quad&\textrm{Karena nilai}\: \: \color{red}m\: \color{blue}\textrm{berada pada}\: \: \color{red}1<m<2\\ &\textrm{maka}\\ &\qquad \color{red}1<m<2\\ &\Leftrightarrow \: \color{red}12<12m<24\\ &\Leftrightarrow \: \color{red}-12>-12m>-24\\ &\Leftrightarrow \: \color{red}9-12>\color{black}9-12m\color{red}>-13\\ &\Leftrightarrow \: \color{red}-3>\color{black}9-12m\color{red}>-13\\ &\Leftrightarrow \: \color{red}-13<\color{black}9-12m\color{red}<-3\\ &\textrm{Ini berarti nilai}\: D\: \: \textrm{negatif, sehingga}\\ &\textrm{berakibat penyebut berupa}\: \: -x^{2}+3x-3m\\ &\color{black}\textrm{adalah wilayah}\: \: \color{red}definit\: negatif\\ 4.\quad&\textrm{Selanjutnya pemfaktoran pertidaksamaan}\\ &\bullet \: \color{black}\textrm{semula}\\ &\quad \displaystyle \frac{x(x+4)}{\underset{\color{red}definit\: negatif}{\underbrace{-x^{2}+3x-3m}}}>0\color{black}\Leftrightarrow \frac{x(x+4)}{-}>0\\ &\bullet \: \color{black}\textrm{akan berubah menjadi}\\ &\qquad \color{black}x(x+4)<0\\ &\qquad\textrm{pembuat nol-nya adalah}:\: \color{red}x(x+4)=0\\ &\qquad\textrm{maka}\: \: \color{red}x=-4\: \: \color{blue}\textrm{atau}\: \color{red}x=0,\: \color{blue}\textrm{sehingga}\\ &\qquad \textrm{interval nilai}\: \color{red}x\color{blue}-\textrm{nya}\: : \color{red}-4<x<0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 65.&\textrm{Jika}\: \: 1<a<2\: ,\: \textrm{maka semua nilai}\\ &x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{-x^{2}+2ax-6}{x^{2}+3x}\leq 0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-3\: \: \textrm{atau}\: \: x>0\\ \textrm{b}.&x<-3\: \: \textrm{atau}\: \: x\geq -2\\ \textrm{c}.&x\leq -2\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{d}.&-3<x<0\\ \textrm{e}.&-2\leq x<0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}1.\quad&\textrm{Diketahui bahwa}:\: \: \color{red}\displaystyle \frac{-x^{2}+2ax-6}{x^{2}+3x}\leq 0\\ &\textrm{untuk membedakan}\: \: a\: \: \textrm{pada persamaan}\\ &\textrm{kuadrat dengan}\: \: a\: \: \textrm{di atas, selanjutnya}\\ &\textrm{kita menuliskan}\: a\: \textrm{di atas dengan}:\: \color{red}m\\ &\color{black}\textrm{karena}\: \: 1<a<2\: \: \textrm{diubah}:1<m<2\\ &\textrm{Perhatikanlah pembilang yang}\\ &\textrm{mengandung bilangan}\: \: \color{red}m\: \: \color{blue}\textrm{yang terletak}\\ & \textrm{pada interval}\: :\: \color{red}1<m<2.\\ 2.\quad&\textrm{Kita cek kondisi pembilangnya dengan}\\ &\textrm{menentukan}\: Diskriminan(D)-\textrm{nya}\\ &\textrm{yaitu}:\\ &\color{red}ax^{2}+bx+c \color{black}\begin{cases} \color{black}a>0\: \&\: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit positif} \\ \color{red}a<0\: \& \: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit negatif} \end{cases}\\ &\textrm{Karena pebilangnya}\: :\: \color{red}-x^{2}+2mx-6,\\ &\textrm{dengan}\: \color{red}a=-1,\: b=2m,\: \&\: c=-6,\: \color{blue}\textrm{maka}\\ &\color{black}D=\color{red}(2m)^{2}-4(-1)(-6)=\color{black}4m^{2}-24\\ 3.\quad&\textrm{Karena nilai}\: \: \color{red}m\: \color{blue}\textrm{berada pada}\: \: \color{red}1<m<2\\ &\textrm{maka}\\ &\qquad \color{red}1<m<2\\ &\Leftrightarrow \: \color{red}1^{2}<m^{2}<2^{2}\color{blue}\Leftrightarrow \color{red}1<m^{2}<4\\ &\Leftrightarrow \: \color{red}4<4m^{2}<16\\ &\Leftrightarrow \: \color{red}4-24<\color{black}4m^{2}-24\color{red}<16-24\\ &\Leftrightarrow \: \color{red}-20<\color{black}4m^{2}-24\color{red}<-8\\ &\textrm{Ini berarti nilai}\: D\: \: \textrm{negatif, sehingga}\\ &\textrm{berakibat pembilangnya berupa}\: \: -x^{2}+2mx-6\\ &\color{black}\textrm{adalah wilayah}\: \: \color{red}definit\: negatif\\ 4.\quad&\textrm{Selanjutnya pemfaktoran pertidaksamaan}\\ &\bullet \: \color{black}\textrm{semula}\\ &\quad \displaystyle \frac{\overset{\color{red}definit\: negatif}{\overbrace{\color{blue}-x^{2}+2mx-6}}}{x(x+3)}\leq 0\color{black}\Leftrightarrow \frac{-}{x(x+3)}\leq 0\\ &\bullet \: \color{black}\textrm{akan berubah menjadi}\\ &\qquad \color{black}x(x+3)> 0\\ &\qquad\textrm{pembuat nol-nya adalah}:\: \color{red}x(x+3)=0\\ &\qquad\textrm{maka}\: \: \color{red}x=-3\: \: \color{blue}\textrm{atau}\: \color{red}x=0,\: \color{blue}\textrm{sehingga}\\ &\qquad \textrm{interval nilai}\: \color{red}x\color{blue}-\textrm{nya}\: : \color{red}x<-3\: \: \textrm{atau}\: \: x>0 \end{aligned} \end{array}$.

Latihan Soal 6 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 46.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{2}{x+1}\leq \left | x \right |\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{b}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: 0\leq x\leq 1 \right \}\\ \textrm{c}.&\left \{ x|x\geq 1 \right \}\\ \color{red}\textrm{d}.&\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{e}.&\left \{ x|-1< x\leq 1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | x \right |\geq \displaystyle \frac{2}{x+1}\quad\quad\quad \color{black}\textrm{berakibat}\\ &\displaystyle \frac{-2}{x+1}\geq x\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{2}{x+1}\\ &\bullet \qquad \color{red}\textrm{bagian}\: \: 1\\ &x\leq \displaystyle \frac{-2}{x+1}\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ &x+\displaystyle \frac{2}{x+1}\leq 0\\ &\displaystyle \frac{x(x+1)+2}{x+1}\leq 0\\ &\displaystyle \frac{x^{2}+x+2}{x+1}\leq 0\Leftrightarrow \displaystyle \frac{\textrm{Definit positif}}{x+1}\leq 0\\ &\begin{aligned}&\textrm{HP}_{1}=\color{black}\left \{x| x< -1,\: x\in \mathbb{R} \right \}\\ &\bullet \qquad \color{red}\textrm{bagian}\: \: 2\\ &x\geq \displaystyle \frac{2}{x+3}\\ &x-\displaystyle \frac{2}{x+1}\geq 0\\ &\displaystyle \frac{x(x+1)-2}{x+1}\geq 0\\ &\displaystyle \frac{x^{2}+x-2}{x+1}\geq 0\\ &\displaystyle \frac{(x+2)(x-1)}{x+1}\geq 0\\ &\textrm{HP}_{2}=\color{black}\left \{x|-2\leq x< -1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \}\\ &\textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}=\color{red}\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 47.&\textrm{Diketahui pertidaksamaan}\: \: \displaystyle \frac{x+10}{x-9}\leq 0\\ &\textrm{dan diberikan beberapa nilai berikut}\\ &(\textrm{i})\quad x=-6\: \, \qquad\qquad (\textrm{iii})\quad x=-14\\ &(\textrm{ii})\, \, \, \: x=-10\qquad\quad\quad (\textrm{iv})\quad x=-18\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\textrm{di atas adalah ditunjukkan oleh}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&(\textrm{i})\: \: \textrm{dan} \: \: (\textrm{ii})\\ \textrm{b}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{c}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{d}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iv})\\ \textrm{e}.&(\textrm{iii})\: \: \textrm{dan}\: \: \: (\textrm{iv}) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x+10}{x-9}&\leq 0\\ \textrm{HP}=&\color{red}\left \{ x|-10\leq x< 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 48.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\leq x< 6\\ \textrm{b}.&2\leq x< 3\\ \color{red}\textrm{c}.&2< x< 3\: \: \textrm{atau}\: \: x>6\\ \textrm{d}.&x<3\: \: \textrm{atau}\: \: 3<x<6\\ \textrm{e}.&x<2\: \: \textrm{atau}\: \: x>3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\\ &\Leftrightarrow \displaystyle \frac{6}{x-3}-\frac{8}{x-2}<0\\ &\Leftrightarrow \displaystyle \frac{6(x-2)-8(x-3)}{(x-3)(x-2)}<0\\ &\Leftrightarrow \displaystyle \frac{6x-8x-12+24}{(x-2)(x-3)}<0\\ &\Leftrightarrow \displaystyle \frac{-2x+12}{(x-2)(x-3)}<0\\ &\Leftrightarrow \displaystyle \frac{2x-12}{(x-2)(x-3)}>0\\ &\Leftrightarrow \displaystyle \frac{2(x-6)}{(x-2)(x-3)}>0\\ &\textrm{HP}=\color{red}\left \{ x|2<x<3\: \: \textrm{atau}\: \: x>6,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 49.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{x^{2}-81}{x^{2}}\geq 0\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x\leq -9\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{b}.&-9\leq x< 0\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{c}.&-9\leq x< 0\: \: \textrm{atau}\: \: 0<x\leq 9\\ \textrm{d}.&-9< x\leq 9\\ \textrm{e}.&x\in \mathbb{R} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{x^{2}-81}{x^{2}}\geq 0\\ &\displaystyle \frac{(x+9)(x-9)}{x^{2}}\geq 0\\ &\textrm{HP}=\color{red}\left \{ x|x\leq -9\: \: \textrm{atau}\: \: x\geq 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}-4}{x+2}>0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x>2\\ \textrm{b}.&-2\leq x< 2\\ \textrm{c}.&x<-2\: \: \textrm{atau}\: \: x>2\\ \textrm{d}.&x<-2\: \: \textrm{atau}\: \: -2< x< 2\\ \textrm{e}.&x\geq -2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{x^{2}-4}{x+2}>0\\ &\displaystyle \frac{(x+2)(x-2)}{(x+2)}>0\\ &(x-2)>0\\ &\color{red}x>2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 51.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}<0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|-9< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{b}.&\left \{ x|-6< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \}\\ \color{red}\textrm{d}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: \displaystyle \frac{5}{2}<x<5,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\left \{ x|x< -9\: \: \textrm{atau}\: \: -6< x< \displaystyle \frac{5}{2}\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}&<0\\ \displaystyle \frac{(x+6)(x-5)}{(x+9)(2x-5)}&<0\\ \color{red}\textrm{Cukup jelas}& \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 52.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{2x+6}{x-4}\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-10<x<4\\ \color{red}\textrm{b}.&-10\leq x<4\\ \textrm{c}.&-4<x\leq 10\\ \textrm{d}.&x\leq -10\: \: \textrm{atau}\: \: x\geq 4\\ \textrm{e}.&x<-10\: \: \textrm{atau}\: \: x\geq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{2x+6}{x-4}&\leq 1\\ \displaystyle \frac{2x+6}{x-4}-1&\leq 0\\ \displaystyle \frac{2x+6-(x-4)}{x-4}&\leq 0\\ \displaystyle \frac{x+10}{x-4}&\leq 0\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 53.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \displaystyle \frac{x^{2}-x}{x+3}\geq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-3\: \: \textrm{atau}\: \: -1\leq x\leq 3\\ \color{red}\textrm{b}.&-3< x\leq -1\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{c}.&-3\leq x\leq 3\\ \textrm{d}.&-3\leq x\leq -1\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{e}.&-3\leq x\leq -1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{x^{2}-x}{x+3}&\geq 1\\ \displaystyle \frac{x^{2}-x}{x+3}&-1\geq 0\\ \displaystyle \frac{x^{2}-x-(x+3)}{x+3}&\geq 0\\ \displaystyle \frac{x^{2}-2x-3}{x+3}&\geq 0\\ \displaystyle \frac{(x-3)(x+1)}{x+3}&\geq 0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 54.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x+2+\displaystyle \frac{1}{x+4}>0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-4\: \: \textrm{atau}\: \: x\geq -3\\ \textrm{b}.&x<-4\: \: \textrm{atau}\: \: x>-3\\ \textrm{c}.&-4\leq x\leq -3\\ \color{red}\textrm{d}.&x>-4\\ \textrm{e}.&-4\leq x\leq -3\: \: \textrm{atau}\: \: x>-3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}x+2+\displaystyle \frac{1}{x+4}&>0\\ \displaystyle \frac{(x+2)(x+4)+1}{(x+4)}&>0\\ \displaystyle \frac{x^{2}+6x+8+1}{x+4}&>0\\ \displaystyle \frac{x^{2}+6x+9}{x+4}&>0\\ \displaystyle \frac{(x+3)^{2}}{(x+4)}&>0\\ x&>-4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 55.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x+3<\displaystyle \frac{x^{2}+6x+11}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|x< -3\displaystyle \frac{2}{3}\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \}\\ \textrm{b}.&\left \{ x|0\leq x\leq 11,\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\left \{ x|x<-11\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \}\\ \textrm{d}.&\left \{ x|x<0\: \: \textrm{atau}\: \: x>11,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\left \{ x|x\leq 11\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}x+3<\displaystyle \frac{x^{2}+6x+11}{x}&\\ x+3-\displaystyle \frac{x^{2}+6x+11}{x}&<0\\ \displaystyle \frac{x(x+3)-\left (x^{2}+6x+11 \right )}{x}&<0\\ \displaystyle \frac{x^{2}+3x-x^{2}-6x-11}{x}&<0\\ \displaystyle \frac{-3x-11}{x}&<0\\ \displaystyle \frac{3x+11}{x}&>0 \end{aligned} \end{array}$.

Latihan Soal 5 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{l}\\ 36.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 3-\left | x \right | \right |<10\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-14\: \: \textrm{atau}\: x>12\\ \textrm{b}.&x<-13\: \: \textrm{atau}\: x>13\\ \textrm{c}.&x<-12\: \: \textrm{atau}\: x>10\\ \textrm{d}.&0<x<10\\ \color{red}\textrm{e}.&-13<0<13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | 3-\left | x \right | \right |< 10\\ &-10< 3-\left | x \right |< 10\\ &-13< -\left | x \right |< 7\\ &-7< \left | x \right |\leq 13,\: \: (\textrm{ingat harga}\: \: \left | x \right |\geq 0)\\ &0\leq \left | x \right |< 13\\ &\textrm{selanjutnya},\\ &\left | x \right |< 13\\ &-13< \: x< 13\\ &\textrm{HP}=\color{red}\left \{ x|\: -13< x< 13,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{l}\\ 37.&\textrm{(UM UGM 05)}\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x^{2}-3 \right |<2x\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<3\\ \textrm{b}.&-3<x<1\\ \color{red}\textrm{c}.&1<x<3\\ \textrm{d}.&-3<x<-1\: \: \textrm{atau}\: \: 1<x<3\\ \textrm{e}.&x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\left | x^{2}-3 \right |&<2x\\ -2x<\left ( x^{2}-3 \right )&<2x\\ \textrm{dipartisi men}&\textrm{jadi dua bagian}\\ \bullet \quad\textrm{pertama}\qquad&\\ (x^{2}-3)&>-2x\\ x^{2}+2x-3&>0\\ (x+3)(x-1)&>0\\ x<-3\: \: \textrm{atau}&\: \: x>1\\ \bullet \quad \textrm{kedua}\qquad\quad&\\ \left ( x^{2}-3 \right )&<2x\\ x^{2}-2x-3&<0\\ (x-3)(x+1)&<0\\ -1<x<3&\\ \color{black}\textrm{ambil yang}&\: \color{black}\textrm{memenuhi keduanya}\\ \textrm{berupa iris}&\textrm{an}\\ \textrm{HP}=&\color{red}\left \{ 1<x<3,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&(\textrm{SPMB 05})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x-2 \right |^{2}<4\left | x-2 \right |+12\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x\in \mathbb{R}|2\leq x\leq 8 \right \}\\ \textrm{b}.&\left \{ x\in \mathbb{R}|4<x< 8 \right \}\\ \color{red}\textrm{c}.&\left \{ x\in \mathbb{R}|-4<x< 8 \right \}\\ \textrm{d}.&\left \{ x\in \mathbb{R}|-2<x<4 \right \}\\ \textrm{e}.&\left \{ x\in \mathbb{R}|2<x<4 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{misalkan}\: \: p&=\left | x-2 \right |,\: \: \textrm{selanjutnya}\\ \left | x-2 \right |^{2}<&\, 4\left | x-2 \right |+12\\ p^{2}<&\, 4p+12\\ p^{2}-&4p-12<0\\ (p-6)&(p+2)<0\\ -2<p&<6,\: \: \color{magenta}\textrm{atau jika dikembalikan}\\ -2<&\left | x-2 \right |<6,\\ &\: \: \color{black}\textrm{ingat, nilanya tidak negatif}\\ 0\leq &\left | x-2 \right |<6\\ -6<&\: x-2<6\\ -4<&\: x<8\\ \textrm{HP}=&\color{red}\left \{ -4<x<8,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Diketahui grafik fungsi}\: \: f(x)=mx^{2}-2mx+m\\ & \textrm{berada di atas grafik fungsi}\\ &g(x)=2x^{2}-3,\: \textrm{maka nilai}\: \: m\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&m>2&&&\textrm{d}.&-6<m<2\\ \textrm{b}.&m>6&\textrm{c}.&2<m<6&\textrm{e}.&m<-6 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}f(x)&=g(x)\\ mx^{2}-2mx+m&=2x^{2}-3\\ mx^{2}-2x^{2}-2mx+m+3&=0\\ \textrm{Supaya grafik}\: \: f(x)\: \textrm{berada }&\color{blue}\textrm{di atasnya}, \\ \textrm{maka}\: \: D=B^{2}-4AC&<0\\ (m-2)x^{2}-2mx+(m+3)&=0\begin{cases} A &=m-2 \\ B &=-2m \\ C &=m+3 \end{cases}\\ B^{2}-4AC&<0\\ (-2m)^{2}-4(m-2)(m+3)&<0\\ 4m^{2}-4\left ( m^{2}+m-6 \right )&<0\\ 4m^{2}-4m^{2}-4m+24&<0\\ -4m+24&<0\\ m-6&>0\\ m&\color{red}>6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 40.&\textrm{Jika}\: \: \color{red}3<x<5\: \: \color{black}\textrm{maka penyelesaian untuk}\\ &\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{lllllll}\\ \textrm{a}.&2x-2&&&\textrm{d}.&-2\\\\ \textrm{b}.&2\quad&\textrm{c}.&8-2x\quad&\textrm{e}.&2x-8\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}\\ &=\sqrt{(x-3)^{2}}-\sqrt{(x-5)^{2}}\\ &=\left | x-3 \right |-\left | x-5 \right |\\ &\textrm{ingat bahwa saat}\: \: \color{red}3<x<5\\ &\textrm{maka}\: \begin{cases} \left | x-3 \right |=(x-3) \\ \left | x-5 \right |=-(x-5) \end{cases},\\ &\color{blue}\textrm{sehingga}\\ &=\left | x-3 \right |-\left | x-5 \right |=(x-3)-\left ( -(x-5) \right )\\ &=x-3+x-5\\ &=\color{red}2x-8 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 41.&\textrm{Jika}\: \: \color{red}1<x<5\: \: \color{black}\textrm{maka penyelesaian untuk}\\ &\sqrt{x^{2}-2x+1}+\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{lllllll}\\ \textrm{a}.&2&&&\textrm{d}.&5\\\\ \textrm{b}.&3\quad&\textrm{c}.&4\quad&\textrm{e}.&6\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\sqrt{x^{2}-2x+1}+\sqrt{x^{2}-10x+25}\\ &=\sqrt{(x-1)^{2}}+\sqrt{(x-5)^{2}}\\ &=\left | x-1 \right |+\left | x-5 \right |\\ &\textrm{ingat bahwa saat}\: \: \color{red}1<x<5\\ &\textrm{maka}\: \begin{cases} \left | x-1 \right |=(x-1) \\ \left | x-5 \right |=-(x-5) \end{cases},\\ &\color{blue}\textrm{sehingga}\\ &=\left | x-1 \right |+\left | x-5 \right |=(x-1)+\left ( -(x-5) \right )\\ &=x-1+5-x\\ &=\color{red}4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 42.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{5}{4x-3} \right |\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\leq x\leq \frac{3}{4}\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{b}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \displaystyle \frac{3}{4}< x\leq 2\\ \textrm{c}.&-\displaystyle \frac{1}{2}\leq x\leq 2,\: \: x\neq \frac{3}{4}\\ \textrm{d}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x>\frac{3}{4}\\ \color{red}\textrm{e}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x\geq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\left | \displaystyle \frac{5}{4x-3} \right |&\leq 1\\ -1\leq \displaystyle \frac{5}{4x-3}&\leq 1,\: \: \color{magenta}\textbf{jika dibalik}\\ -1\geq \displaystyle \frac{4x-3}{5}&\geq 1,\: \: \color{magenta}\textbf{bentuk ini tidak}\\ \color{magenta}\textbf{dibolehkan}&\: \color{magenta}\textbf{maka perlu diubah menjadi}\\ -1\geq \displaystyle \frac{4x-3}{5}\: \: \textrm{atau}&\: \: \displaystyle \frac{4x-3}{5}\geq 1,\: \: \color{black}\textrm{selanjutnya}\\ \bullet \quad \textrm{bagian}&\: 1\\ -1&\geq \displaystyle \frac{4x-3}{5}\Leftrightarrow \frac{4x-3}{5}\leq -1\\ 4x-3&\leq -5\\ 4x&\leq -2\\ x&\leq -\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{4x-3}{5}&\geq 1\\ 4x-3&\geq 5\\ 4x&\geq 8\\ x&\geq 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 43.&(\textrm{UMPTN 95})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2}{2x-1} \right |> 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x> 2\\ \textrm{b}.&x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{c}.&x<-1\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{d}.&-1<x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{e}.&x<-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{semua opsi bukan jawaban}\\ &\textbf{Berikut pembahasannya}\\ &\begin{aligned}\left | \displaystyle \frac{2}{2x-1} \right |&> 1\\ -1>\displaystyle \frac{2}{2x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2}{2x-1}>1,\: \color{magenta}\textbf{dibalik}\\ -1<\displaystyle \frac{2x-1}{2}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x-1}{2}<1\\ \bullet \quad \textrm{bagian}&\: 1\\ \displaystyle \frac{2x-1}{2}&>-1\\ 2x-1&>-2\\ 2x&>-1\\ x&>-\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{2x-1}{2}&<1\\ 2x-1&<2\\ 2x&<3\\ x&<\displaystyle \frac{3}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 44.&(\textrm{UMPTN 00})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2x+7}{x-1} \right |\geq 1\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-2\leq x\leq 8\\ \textrm{b}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -2\\ \textrm{c}.&-8\leq x< 1\: \: \textrm{atau}\: \: x>1\\ \textrm{d}.&-2\leq x< 1\: \: \textrm{atau}\: \: 1< x\leq 8\\ \color{red}\textrm{e}.&x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\left | \displaystyle \frac{2x+7}{x-1} \right |&\geq 1\\ -1\geq \displaystyle \frac{2x+7}{x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x+7}{x-1}\geq 1\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{2x+7}{x-1}&\leq -1\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{2x+7}{x-1}&+1\leq 0\\ &\displaystyle \frac{2x+7+(x-1)}{x-1}\leq 0\\ \displaystyle \frac{3x+6}{x-1}&\leq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| -2\leq x< 1,\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{2x+7}{x-1}&\geq 1\\ \displaystyle \frac{2x+7}{x-1}&-1\geq 0\\ &\displaystyle \frac{2x+7-(x-1)}{x-1}\geq 0\\ \displaystyle \frac{x+8}{x-1}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>1,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x> 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{x-2}{x+3} \right |\leq 2\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-8\leq x< -3\\ \textrm{b}.&-8\leq x< -1\\ \textrm{c}.&-4\leq x< -3\\ \color{red}\textrm{d}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3}\\ \textrm{e}.&x\leq -4\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left | \displaystyle \frac{x-2}{x+3} \right |&\leq 2\\ -2\leq \displaystyle \frac{x-2}{x+3}&\: \: \textrm{atau}\: \: \displaystyle \frac{x+2}{x+3}\leq 2\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{x-2}{x+3}&\geq -2\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{x-2}{x+3}&+2\geq 0\\ &\displaystyle \frac{x-2+2(x+3)}{x+3}\geq 0\\ \displaystyle \frac{3x+4}{x+3}&\geq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| x< -3\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{x-2}{x+3}&\leq 2\\ \displaystyle \frac{x-2}{x+3}&-2\leq 0\\ &\displaystyle \frac{x-2-2(x+3)}{x+3}\leq 0\\ \displaystyle \frac{-x-8}{x+3}&\leq 0,\: \: \color{magenta}\textbf{koefisien \textit{x} negatif}\\ \displaystyle \frac{x+8}{x+3}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>-3,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

Latihan Soal 4 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 26.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & (x+3)(x-1)\geq (x-1)\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&1\leq x\leq 3&\\ \textrm{b}.&x\leq -2\: \: \textrm{atau}\: \: x\geq 1\\ \textrm{c}.&-3\leq x\leq -1\\ \textrm{d}.&-2\geq x\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{e}.&-1\geq x\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}(x+3)(x-1)&\geq (x-1)\\ (x+3)(x-1)-(x-1)&\geq 0\\ (x-1)\left ( (x+3)-1 \right )&\geq 0\\ (x-1)(x+2)&\geq 0\\  \end{aligned}\\ &\textrm{Sehingga solusinya adalah:}\\ &\color{red}x\leq -2\: \: \color{black}\textrm{atau}\: \: \color{red}x\geq 1 \end{array}$.

$\begin{array}{ll}\\ 27.&\textrm{Himpunan penyelesaian pertidaksamaan}\\ & \left | x+3 \right |<2\left | x-4 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&\left \{ x|x<-\displaystyle \frac{5}{3} \right \}\\ \textrm{b}.&\left \{ x|\: \displaystyle \frac{5}{3}<x<-11 \right \}\\ \textrm{c}.&\left \{ x|x\geq -11 \right \} \\ \textrm{d}.&\left \{ x|x<\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x>11 \right \}\\ \textrm{e}.&\left \{ x|x>-\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x<-11 \right \} \\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | x+3 \right |<2\left | x-4 \right |\\ &\left ( x+3 \right )^{2}<2^{2}\left ( x-4 \right )^{2}\\ & \textrm{dikuadratkan masing-masing ruas}\\ &x^{2}+6x+9<4\left ( x^{2}-8x+16 \right )\\ &x^{2}-4x^{2}+6x+32x+9-64<0\\ &-3x^{2}+38x-55<0\\ &3x^{2}-38x+55>0\\ &\left ( 3x-5 \right )\left (x -11 \right )>0\\\\ &\textrm{Berikut untuk}\: \textrm{garis bilangannya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Himpunan penyelesaian pertidaksamaan}\\ & \left | x^{2}+5x \right |\leq 6\: \: \textrm{adalah... .}\\ &\textrm{a}.\quad \left \{ x|-6\leq x\leq 1 \right \}\\ &\textrm{b}.\quad \left \{ x|-3\leq x\leq -2 \right \}\\ &\textrm{c}.\quad \left \{ x|-6\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 1\right \}\\ &\textrm{d}.\quad \left \{ x|-6\leq x\leq -5 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\ &\textrm{e}.\quad \left \{ x|-5\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diketahui bahwa}\\ &\\ \left | x^{2}+5x \right |&\leq 6\\ -6\leq x^{2}+5x&\leq 6\\ & \end{aligned}\\ &\begin{array}{|c|c|}\hline \begin{aligned}&-6\leq x^{2}+5x\\ &x^{2}+5x+6\geq 0\\ &(x+3)(x+2)\geq 0 \end{aligned}&\begin{aligned}&x^{2}+5x\leq 6\\ &x^{2}+5x-6\leq 0\\ &(x+6)(x-1)\leq 0 \end{aligned}\\\hline \textbf{Lihat Gambar 1}&\textbf{Lihat Gambar 2}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \left | \displaystyle \frac{1}{2}x+6 \right |\geq 9\\ & \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-12< x< 6\\ \textrm{b}.&-30\leq x\leq 6\\ \color{red}\textrm{c}.&x\geq 6\: \: \textrm{atau}\: x\leq -30\\ \textrm{d}.&x<6\: \: \textrm{atau}\: \: x<-30\\ \textrm{e}.&x\leq 6\: \: \textrm{atau}\: \: x\geq -30\end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | \displaystyle \frac{1}{2}x+6 \right |\geq 9\\ &\displaystyle \frac{1}{2}x+6\leq -9\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}x+6\geq 9\\ &\displaystyle \frac{1}{2}x\leq -9-6\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}x\geq 9-6\\ &\displaystyle \frac{1}{2}x\leq -15\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}x\geq 3\\ &\color{red}x\leq -30\: \: \color{black}\textrm{atau}\: \: \color{red}x\geq 6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &3\left | x+1 \right |\leq \left | x-2 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\leq x\leq -\frac{1}{4}\\ \color{red}\textrm{b}.&-\displaystyle \frac{5}{2}\leq x\leq \frac{5}{2}\\ \textrm{c}.&x\leq \displaystyle \frac{1}{4}\: \: \textrm{atau}\: x\geq \frac{5}{2}\\ \textrm{d}.&x\leq -\displaystyle \frac{5}{2}\: \: \textrm{atau}\: x\geq \frac{1}{4}\\ \textrm{e}.&x\leq -\displaystyle \frac{5}{2}\: \: \textrm{atau}\: x\geq -\frac{1}{4}\end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&3\left | x+1 \right |\leq \left | x-2 \right |\\ &\left (3\left | x+1 \right | \right )^{2}\leq \left (\left | x-2 \right | \right )^{2}\\ &\left ( 3x+3 \right )^{2}\leq \left (x-2 \right )^{2}\\ &(3x+3+(x-2))(3x+3-(x-2))\leq 0\\ &(4x+1)(2x+5)\leq 0\\ &\textrm{HP}=\color{red}\left \{ x|-\displaystyle \frac{5}{2}\leq x\leq -\frac{1}{4},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{l}\\ 31.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x-3 \right |<3\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<3\\ \textrm{b}.&-3<x<3\\ \textrm{c}.&x<-3\: \: \textrm{atau}\: x<3\\ \color{red}\textrm{d}.&x>0\: \: \textrm{atau}\: x<6\\ \textrm{e}.&x<0\: \: \textrm{atau}\: x<6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | x-3 \right |<3\\ &-3<(x-3)<3\\ &-3+3<x<3+3\\ &\color{red}0<x<6 \end{aligned} \end{array}$

$\begin{array}{l}\\ 32.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x+4 \right |>8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x>-8\\ \textrm{b}.&x<4\: \: \textrm{atau}\: x>12\\ \textrm{c}.&x>4\: \: \textrm{atau}\: x>-12\\ \textrm{d}.&x<-4\: \: \textrm{atau}\: x<6\\ \color{red}\textrm{e}.&x>4\: \: \textrm{atau}\: x<-12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | x+4 \right |>8\\ &(x+4)<-8\: \: \textrm{atau}\: \: (x+4)>8\\ &\color{red}x<-12\: \: \color{black}\textrm{atau}\: \: \color{red}x>4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Himpunan penyelesaian dari pertidaksamaan}\\ &\left | \displaystyle \frac{x+1}{2} \right |>\left | \displaystyle \frac{x-2}{3} \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{HP}=\left \{ x|-7<x<\displaystyle \frac{1}{5},\: x\in \mathbb{R} \right \}\\ \color{red}\textrm{b}.&\textrm{HP}=\left \{ x|x<-7\: \: \textrm{atau}\: \: x>\displaystyle \frac{1}{5},\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\textrm{HP}=\left \{ x|x>-7,\: x\in \mathbb{R} \right \}\\ \textrm{d}.&\textrm{HP}=\left \{ x|-1<x<2,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\textrm{HP}=\left \{ x|x<-1\: \: \textrm{atau}\: \: x>2,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\left | \displaystyle \frac{x+1}{2} \right |>\left | \displaystyle \frac{x-2}{3} \right |\\ &\left ( \displaystyle \frac{x+1}{2} \right )^{2}>\left ( \displaystyle \frac{x-2}{3} \right )^{2}\\ &\left ( \displaystyle \frac{x+1}{2}+\frac{x-2}{3} \right )\left ( \displaystyle \frac{x+1}{2}-\displaystyle \frac{x-2}{3} \right )>0\\ &\left ( \displaystyle \frac{3(x+1)+2(x-2)}{6} \right )\left ( \displaystyle \frac{3(x+1)-2(x-2)}{6} \right )>0\\ &\left ( \displaystyle \frac{5x-1}{6} \right )\left ( \displaystyle \frac{x+7}{6} \right )>0\\ &\textrm{HP}=\color{red}\left \{ x|x<-7\: \: \color{black}\textrm{atau}\: \: \color{red}x>\displaystyle \frac{1}{5},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 34.&\textrm{Penyelesaian dari pertidaksamaan}\\ &\left | \displaystyle \frac{3-2x}{-5} \right |>5\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-11\: \: \textrm{atau}\: x>14\\ \textrm{b}.&x<-14\: \: \textrm{atau}\: x>11\\ \textrm{c}.&11<x<14\\ \textrm{d}.&-14<x<-11\\ \textrm{e}.&x>14 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\left | \displaystyle \frac{3-2x}{-5} \right |>5\\ &\displaystyle \frac{3-2x}{-5}<-5\: \: \textrm{atau}\: \: \displaystyle \frac{3-2x}{-5}>5\\ &\displaystyle \frac{2x-3}{5}>5\: \: \textrm{atau}\: \: \displaystyle \frac{2x-3}{5}<-5\\ &2x-3>25\: \: \textrm{atau}\: \: 2x-3<-25\\ &2x>25+3\: \: \textrm{atau}\: \: 2x<-25+3\\ &x>14\: \: \textrm{atau}\: \: x<-11,\\ &\textrm{dapat juga dituliskan}\\ &\color{red}x<-11\: \: \color{black}\textrm{atau}\: \: \color{red}x>14 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 2-2\left | x+1 \right | \right |>4\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-4\: \: \textrm{atau}\: x>2\\ \textrm{b}.&x<-3\: \: \textrm{atau}\: x>1\\ \textrm{c}.&x<-2\: \: \textrm{atau}\: x>0\\ \textrm{d}.&x<-1\: \: \textrm{atau}\: x>3\\ \textrm{e}.&x<0\: \: \textrm{atau}\: x>4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\left | 2-2\left | x+1 \right | \right |>4\\ &2-2\left | x+1 \right |<-4\: \: \textrm{atau}\: \: 2-2\left | x+1 \right |>4\\ &-2\left | x+1 \right |<-6\: \: \textrm{atau}\: \: -2\left | x+1 \right |>2\\ &\left | x+1 \right |>3\: \: \textrm{atau}\: \: \left | x+1 \right |<-1\\ &\left\{\begin{matrix} (x+1)<-3\\ (x+1)>3 \end{matrix}\right.\: \: \textrm{atau}\: \: \left\{\begin{matrix} \left | x+1 \right |<-1\\ \color{red}\textbf{tak mungkin} \end{matrix}\right.\\ &\textrm{Selanjutnya}\: \textrm{akan didapatkan}\\ &\color{red}x<-4\: \: \color{black}\textrm{atau}\: \: \color{red}x>2 \end{aligned} \end{array}$

Latihan Soal 3 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 16.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | 3x-(4x-7) \right |=6\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -13,-1 \right \}&&&\textrm{d}.&\left \{ -13,1 \right \}\\ \textrm{b}.&\left \{ -1,13 \right \}&\textrm{c}.&\left \{ 1,13 \right \}&\textrm{e}.&\left \{ 13 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\left | 3x-(4x-7) \right |&=6\\ \left | 3x-4x+7 \right |&=6\\ \left | -x+7 \right |&=6\\ \left ( -x+7 \right )&=\pm 6\\ -x+7&=\begin{cases} 6 \Leftrightarrow -x=6-7\Leftrightarrow x=\color{red}1\\\\ -6\Leftrightarrow -x=-6-7\Leftrightarrow x=\color{red}13 \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | x-1 \right |=2x+1\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ -2,0 \right \}&\textrm{c}.&\left \{ -1 \right \}&\textrm{e}.&\left \{ 0 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | x-1 \right |=2x+1\\ &(x-1)=\pm (2x+1)\\ &(x-1)= \begin{cases} +(2x+1) \\\\ -(2x+1) \end{cases}\\ &\begin{aligned}&\\ \color{blue}\textrm{Syarat}&:\\ (x-1)&\begin{cases} x-1\geq 0 \Leftrightarrow x\geq 1\\\\ x-1<0 \Leftrightarrow x<1 \end{cases}\\ \end{aligned}\\ &\begin{array}{|c|c|}\hline x\geq 1&x< 1\\\hline \textrm{Proses}&\textrm{Proses}\\\hline \begin{aligned}(x-1)&=+(2x+1)\\ x-2x&=1+1\\ -x&=2\\ x&=-2 \end{aligned}&\begin{aligned}(x-1)&=-(2x+1)\\ x+2x&=-1+1\\ 3x&=0\\ x&=0 \end{aligned}\\\hline \textrm{tidak memenuhi}&\color{red}\textbf{memenuhi}\\\hline \end{array} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | 3x+1 \right |=2x+9\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ 8 \right \}&&\textrm{e}.&\textrm{setiap bilangan real}\\ \textrm{c}.&\left \{ -2,8 \right \}& \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | 3x+1 \right |=2x+9\\ &(3x+1)=\pm (2x+9)\\ &(3x+1)= \begin{cases} +(2x+9) \\\\ -(2x+9) \end{cases}\\ \end{aligned}\\ &\begin{aligned} \color{blue}\textrm{Syarat}\: \: &:\\ (3x+1)&\begin{cases} 3x+1\geq 0 \Leftrightarrow x\geq -\displaystyle \frac{1}{3}\\\\ 3x+1<0 \Leftrightarrow x<-\displaystyle \frac{1}{3} \end{cases}\\ & \end{aligned} \\ &\begin{array}{|c|c|}\hline x\geq -\displaystyle \frac{1}{3}&x< -\displaystyle \frac{1}{3}\\\hline \textrm{Proses}&\textrm{Proses}\\\hline \begin{aligned}(3x+1)&=+(2x+9)\\ 3x-2x&=9-1\\ x&=8\\ & \end{aligned}&\begin{aligned}(3x+1)&=-(2x+9)\\ 3x+2x&=-9-1\\ 5x&=-10\\ x&=-2 \end{aligned}\\\hline \color{red}\textbf{memenuhi}&\color{red}\textbf{memenuhi}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Jumlah akar-akar dari}\: \: x^{2}+\left | x \right |-6=0\\ & \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-1\\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&4\\\\ &&&(\textbf{Entrance Examination}) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}x^{2}+\left | x \right |-6&=0\\ (\left | x \right |+3)(\left | x \right |-2)&=0\\ \left | x \right |+3=0\quad \textrm{atau}\quad \left | x \right |-2&=0\\ \left | x \right |=-3\: (\textbf{tm})\quad \textrm{atau}\quad \left | x \right |&=2\: (\textbf{mm})\\ \end{aligned} \\ &\textrm{Selanjutnya}\\ &\begin{aligned} x&=\pm 2\begin{cases} x_{1}&=2 \\ x_{2} &=-2 \end{cases}\\ &\textrm{untuk jumlah}\: \textrm{dari akar-akarnya adalah}:\\ &x_{1}+x_{2}=2+(-2)\\ &=\color{red}0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Penyelesaian pertidaksamaan}\: \: x^{2}+\left | x \right |-6\leq 0\\ & \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-2\leq x< 0\\ \textrm{b}.&0\leq x\leq 2\\ \textrm{c}.&-2\leq x\leq 2\\ \textrm{d}.&-3\leq x\leq 2\\ \textrm{e}.&-2\leq x\leq 3\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\\\ &\color{blue}\textbf{Alternatif 1}\\ &\begin{aligned}&\textrm{Proses penyelesaian dipecah jadi 2 bagian}\\ &\textrm{yaitu}:\begin{cases} x & \geq 0 \\ x & <0 \end{cases}\\ &\textrm{Diketahui pertidaksamaan}:\: x^{2}+\left | x \right |-6\leq 0\\ &\begin{array}{|c|c|}\hline (1)&(2)\\\hline x\geq 0&x<0\\\hline \textrm{maka}\: \: \left | x \right |=x&\textrm{maka}\: \: \left | x \right |=-x\\\hline \begin{aligned}&x^{2}+(x)-6\leq 0\\ &x^{2}+x-6\leq 0\\ &(x+3)(x-2)\leq 0\\ &\color{red}\textrm{Selesaian}:\\ &-3\leq x\leq 2\\ &\textrm{karena}\quad x\geq 0,\\ &\textrm{maka penyelesaian}\\ &\textrm{menjadi}\\ &\color{blue}0\leq x\leq 2 \end{aligned}&\begin{aligned}&x^{2}+(-x)-6\leq 0\\ &x^{2}-x-6\leq 0\\ &(x+2)(x-3)\leq 0\\ &\color{red}\textrm{Selesaian}:\\ &-2\leq x\leq 3\\ &\textrm{karena}\quad x<0,\\ &\textrm{maka penyelesaian}\\ &\textrm{menjadi}\\ &\color{blue}-2\leq x<0 \end{aligned}\\\hline \end{array}\\ &\textrm{Gabungan dari penyelesaian (1) dan (2)}\\ &\textrm{adalah}:\quad \color{red}-2\leq x\leq 2 \end{aligned}\\\\ &\color{blue}\textbf{Alternatif 2}\\ &\begin{aligned}&\textrm{Diketahui pertidaksamaan}:\: x^{2}+\left | x \right |-6\leq 0\\ &\textrm{dan perlu diingat pula bahwa}:\quad \color{red}\left | x \right |\geq 0\\ &\textrm{diubah menjadi}:\quad \left | x \right |^{2}+\left | x \right |-6\leq 0\\ &\Leftrightarrow (\left | x \right |+3)(\left | x \right |-2)\leq 0\\ &\Leftrightarrow -3\leq \left | x \right |\leq 2\\ &\textrm{karena}\quad \left | x \right |\geq 0,\: \textrm{maka}\\ &0\leq \left | x \right |\leq 2\Rightarrow \left | x \right |\leq 2\\ &\textrm{Sehingga penyelesaian menjadi}\\ &\color{red}-2\leq x\leq 2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 21.&\textrm{Seluruh bilangan bilangan real}\: \: x\\ &\textrm{yang jaraknya terhadap 3 kurang dari 1 }\\ &\textrm{adalah}\: ... .\\ &\begin{array}{lllllll}\\ \textrm{a}.&3<x<4&\\ \textrm{b}.&2<x<3 \\ \textrm{c}.&2<x<4 \\ \textrm{d}.&3<x<5\\ \textrm{e}.&1<x<3\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Seluruh }\textrm{bilangan bilangan real}\: \: x\\ &\textrm{yang jaraknya terhadap 3 kurang dari 1, }\\ &\textrm{maksudnya adalah}:\\ &\left | x-3 \right |<1\\ &\Leftrightarrow -1<x-3<1\\ &\Leftrightarrow -1+\textbf{(3)}<x-3+\textbf{(3)}<1+\textbf{(3)}\\ &\Leftrightarrow \color{red}2<x<4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Pernyataan berikut yang tepat adalah}\: ... .\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{b}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<1\\ \textrm{c}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & -3<m<-2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{d}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & 2<m<3\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{e}.&\textrm{pilihan jawaban baik a, b, c,}\\ &\textrm{maupun d tidak ada yang benar} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikanlah opsi}\: \: \textbf{a}\: ,\\ \left | m \right |&<2\\ -2<m&<2\\ \textrm{sehing}&\textrm{ga untuk nilai}\: \: m\in \mathbb{R}\: \: \textrm{pada rentang}\\ -1<m&<2\: \: \: \: \textrm{akan memenuhi semua} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\\ & \left | 2x-9 \right |< 3\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-3\leq x\leq 6&\\ \textrm{b}.&-3<x<6\\ \textrm{c}.&3<x<6\\ \textrm{d}.&3\leq x\leq 6\\ \textrm{e}.&-3<x<-6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | 2x-9 \right |< 3\\ &\Leftrightarrow -3<2x-9<3\\ &-3+(9)<2x-9+(9)<3+(9)\\ &\textnormal{masing-masing ditambah 9}\\ &\textnormal{dan akan menjadi bentuk}\\ &6<2x<12\\ &\color{blue}6.\left ( \displaystyle \frac{1}{2} \right )<2x.\left ( \displaystyle \frac{1}{2} \right )<12.\left ( \displaystyle \frac{1}{2} \right )\\ &\textnormal{masing-masing dikali}\: \: \displaystyle \frac{1}{2}\\ &\textnormal{dan akan berubah menjadi bentuk}\\ &\color{red}3<x<6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\\ & \left | 3x+5 \right |\geq 19\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&x\leq -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x\geq 8&\\\\ \textrm{b}.&x<-8\: \: \textrm{atau}\: \: x>\displaystyle \frac{14}{3}\\\\ \textrm{c}.&x\leq -8\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{14}{3}\\\\ \textrm{d}.&x< -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x> 8\\\\ \textrm{e}.&x\leq 8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{14}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | 3x+5 \right |\geq 19\\\\ &(\ast )-19\geq 3x+5\quad \textrm{atau}&(\ast \ast )\: \: 3x+5\geq 19\\ &-19-5\geq 3x\quad \textrm{atau}&3x\geq 19-5\\ &-\displaystyle \frac{24}{3}\geq x\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ &-8\geq x\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ &x\leq \color{red}-8\quad \color{black}\textrm{atau}&x\color{red}\geq \displaystyle \frac{14}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi}\\ & 25-\left | 10x+5 \right |\geq \left | 40x-20 \right |\: \: \textrm{adalah}... .\\\\ &\qquad (\textbf{NUS Entrance Examination A level})\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}:\\ &\begin{aligned}&25-\left | 10x+5 \right |\geq \left | 40x-20 \right |\\ &25-5\left | 2x+1 \right |\geq 20\left | 2x-1 \right |\\ &5-\left | 2x+1 \right |\geq 4\left | 2x-1 \right |\\ &\color{red}\textrm{ilustrasinya}\quad \color{black}\begin{array}{llllllllll} &&&&&&\\\hline &&-\frac{1}{2}&&&\frac{1}{2}&& \end{array}\\ &\textrm{dan berikut}\: \textrm{pembagian wilayahnya}\\ &\begin{array}{|c|c|c|}\hline -\infty < x< -\displaystyle \frac{1}{2}&-\displaystyle \frac{1}{2}\leq x< \frac{1}{2}&\displaystyle \frac{1}{2}\leq x< \infty \\\hline \begin{cases} \left | 2x+1 \right | &=-(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=+(2x-1) \end{cases}\\\hline \end{array}\\ &\textrm{Selanjutnya adalah} \end{aligned} \end{array}$

$.\qquad\begin{array}{|c|c|c|}\hline \begin{aligned}\: -\infty < x<& -\frac{1}{2}\: ,\\ 25-\left | 10x+5 \right |&\geq \left | 40x-20 \right |\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(-(2x+1))&\geq 4(-(2x-1))\\ 5+2x+1&\geq -8x+4\\ 10x&\geq -2\\ x&\geq -\frac{2}{10}\quad (\textbf{tm})\\ & \end{aligned}&\begin{aligned} \: -\frac{1}{2} \leq x<& \frac{1}{2}\: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1)&\geq 4(-(2x-1))\\ 5-2x-1&\geq -8x+4\\ 6x&\geq 0\\ x&\geq 0\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ 0\leq x< \frac{1}{2} \right \} \end{aligned} &\begin{aligned}\: \frac{1}{2}\leq x< &\infty \: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1))&\geq 4(2x-1)\\ 5-2x-1&\geq 8x-4\\ -10x&\geq -8\\ x&\leq \frac{8}{10}\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ \frac{1}{2}\leq x\leq \frac{4}{5} \right \} \end{aligned}\\\hline \end{array}$

$.\qquad\begin{aligned}&\\ &\textrm{Sehingga yang memenuhi}\: \textrm{adalah}:\\ &=\color{red}\left \{ 0\leq x\leq \displaystyle \frac{4}{5} \right \} \end{aligned}$.

sumber soal di sini dan di sini

Latihan Soal 2 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 11.&\textrm{Perhatikanlah ilustrasi grafik di bawah ini}\\ &\begin{aligned}\end{aligned}\end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Persamaan yang memenuhi rumus tersebut }\\ &\textrm{adalah}\: ... .\\ &\begin{array}{llll}\\ \textrm{a}.&y=\left | -x-2 \right |\\ \textrm{b}.&y=-2x-4\\ \textrm{c}.&y=-\left | 2x-4 \right |\\ \textrm{d}.&y=\left | -2x-4 \right |\\ \textrm{e}.&y=\left | -2x+4 \right | \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Dengan cara substitusi langsung kita}\\ & \textrm{akan mendapatkan}\\ &\bullet \quad \textrm{untuk}\: \: x=4\: \: \textrm{menyebabkan nilai}\: \: y=4\\ &\qquad \textrm{dan sampai langkah di sini hanya ada }\\ &\qquad \textrm{1 persamaan yang memenuhi yaitu}:\\ &\qquad y=\color{red}\left | -2x+4 \right | \end{aligned} \end{array}$.

$\begin{array}{lll}\\ 12.&\textrm{Gambarlah garfik untuk persamaan}\\ & \left | x \right |+\left | y \right |=4\\\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Dari soal diketahui}\\ &\left | x \right |+\left | y \right |=4\\ &\textrm{maka untuk} \end{aligned}\\&\begin{array}{|cc|cc|}\hline x> 0\: ,\: y> 0&&&x> 0\: ,\: y<0\\\hline x+y=4&&&x+(-y)=4\\ &&&\\\hline &&&\\ x<0\: ,\: y> 0&&&x<0\: ,\: y<0\\\hline (-x)+y=4&&&(-x)+(-y)=4\\\hline \end{array} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Perhatikanlah ilustrasinya grafik di bawah ini}\\ &\begin{aligned}\end{aligned}\end{array}$.

$\begin{array}{lll}\\ 13.&\textrm{Tentukanlah nilai}\: \: x\: \: \textrm{dan}\: \: y\\ & \textrm{yang memenuhi}\: \: x+\left | x \right |+y=5\\ & \textrm{dan}\: \: x+\left | y \right |-y=10\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\\ &\begin{aligned} &\left | x \right |+x+y=5\\ &\: \: \qquad \textrm{dan}\\ &x+\left | y \right |-y=10\\ & \end{aligned} \end{aligned} \end{array}$.

$.\: \quad\begin{aligned}\begin{aligned}&\textrm{untuk}:\: \: x> 0\: ,\: y> 0\quad \textbf{(kuadran I)}\\ &\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5 \\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\: (\textrm{ada}) \end{cases}\\ &\textrm{untuk}:\: \: x<0\: ,\: y> 0\quad \textbf{(kuadran II)}\\ &\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\\ &\textbf{(tidak memenuhi)} \end{cases} \end{aligned} \end{aligned}$.

$.\: \quad\begin{aligned}\begin{aligned}&\textrm{untuk}:\: \: x<0\: ,\: y<0\quad \textbf{(kuadran III)}\\ &\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\\ & \textbf{(tidak memenuhi)}\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases}\\ &\textrm{untuk}:\: \: x> 0\: ,\: y<0\quad \textbf{(kuadran IV)}\\ &\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5\\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases} \end{aligned} \end{aligned}$.

$.\quad\begin{array}{ll}\\ &\textrm{Sebagai ilustrasinya, berikut grafiknya}\\ &\begin{aligned}\end{aligned}\end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Gambarlah grafik fungsi mutlak berikut}\\ &\textrm{a}.\quad y=\left | x-2 \right |\\ &\textrm{b}.\quad y=-\left | x-2 \right |\\ &\textrm{c}.\quad y=2+\left | x-2 \right |\\ &\textrm{d}.\quad y=2-\left | x-2 \right |\\ &\textrm{e}.\quad y=\left | 2+\left | x-2 \right | \right |\\ &\textrm{f}.\quad y=\left | 2-\left | x-2 \right | \right | \end{array}$.

$.\quad\begin{array}{ll}\\ &\textbf{Jawab}\quad :\\ &\textrm{Berikut ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{a} \end{array}$.

$.\quad\begin{array}{ll}\\  &\textrm{Dan berikut ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{b} \end{array}$.

$.\quad\begin{array}{ll}\\  &\textrm{Dan berikut pula ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{c} \end{array}$.

$.\quad\begin{array}{ll}\\  &\textrm{Dan berikut pula ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{d} \end{array}$.

$.\quad\begin{array}{ll}\\  &\textrm{Dan sebagai ilustrasinya}\\ &\textrm{lihat soal no}\: \: \textbf{c} \end{array}$.

$.\quad\begin{array}{ll}\\  &\textrm{Dan terakhir berikut ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{f} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | 3x-(4x-7) \right |=6\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -13,-1 \right \}&&&\textrm{d}.&\left \{ -13,1 \right \}\\ \textrm{b}.&\left \{ -1,13 \right \}&\textrm{c}.&\left \{ 1,13 \right \}&\textrm{e}.&\left \{ 13 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\left | 3x-(4x-7) \right |&=6\\ \left | 3x-4x+7 \right |&=6\\ \left | -x+7 \right |&=6\\ \left ( -x+7 \right )&=\pm 6\\ -x+7&=\begin{cases} 6 \Leftrightarrow -x=6-7\Leftrightarrow x=\color{red}1\\\\ -6\Leftrightarrow -x=-6-7\Leftrightarrow x=\color{red}13 \end{cases} \end{aligned} \end{array}$.

Latihan Soal 1 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 1.&\textrm{Nilai untuk}\: \: -\left | -2021 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-2021\quad &&&\textrm{d}.&-2021^{-1}\\ \textrm{b}.&2021&\textrm{c}.&2021^{-1}\quad &\textrm{e}.&2021^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{red}\begin{aligned}-\left | -2021 \right |&=-\left ( 2021 \right )=-2021 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Nilai untuk}\: \: \left | -4 \right |-\left | -6^{2}\times 2 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-68\quad &&&\textrm{d}.&68\\ \textrm{b}.&-40&\textrm{c}.&40\quad &\textrm{e}.&76 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\left | -4 \right |-\left | -6^{2}\times 2 \right |&=\left ( 4 \right )-\left | -36\times 2 \right |\\ &=4-\left | -72 \right |\\ &=4-\left ( 72 \right )\\ &=\color{red}-68 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Nilai untuk}\: \: (-2022)-\left | -(-2021) \right |-\left | -3^{2} \right |^{2}=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-3960\quad &&&\textrm{d}.&-4068\\ \textrm{b}.&-4038&\textrm{c}.&-4050\quad &\textrm{e}.&-4124 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&(-2022)-\left | -(-2021) \right |-\left | -3^{2} \right |^{2}\\ &=-2022-2021-\left | -9 \right |^{2}\\ &=-4043-9^{2}\\ &=-4043-81\\ &=-4124 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Nilai untuk}\: \: \left | -4 \right |^{\left | -2 \right |}-\left | -2 \right |^{\left | -4 \right |}=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-32\quad &&&\textrm{d}.&16\\ \textrm{b}.&-16&\textrm{c}.&0\quad &\textrm{e}.&32 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\left | -4 \right |^{\left | -2 \right |}-\left | -2 \right |^{\left | -4 \right |}&=4^{2}-2^{4}\\ &=16-16\\ &=\color{red}0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Nilai untuk}\\ & \left | 1-2^{2} \right |+\left | 2^{2}-3^{2} \right |+\left | 3^{2}-4^{2} \right |+\cdots +\left | 2020^{2}-2021^{2} \right |...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-2021^{2}\quad &&&\textrm{d}.&2021^{2}-1\\ \textrm{b}.&1-2021^{2}&\textrm{c}.&-2021\quad &\textrm{e}.&2021^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | 1-2^{2} \right |+\left | 2^{2}-3^{2} \right |+\left | 3^{2}-4^{2} \right |+\cdots +\left | 2020^{2}-2021^{2} \right |\\ &=\left ( 2^{2}-1 \right )+\left ( 3^{2}-2^{2} \right )+\left ( 4^{2}-3^{2} \right )+\cdots +\left ( 2021^{2}-2020^{2} \right )\\ &=-1+2021^{2}\\ &=\color{red}2021^{2}-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Nilai untuk}\\ & \left | \displaystyle \frac{1}{2}-1 \right |\times \left | \displaystyle \frac{1}{3}-1 \right |\times \left | \displaystyle \frac{1}{4}-1 \right |\times \left | \displaystyle \frac{1}{5}-1 \right |\times \cdots \times \left | \displaystyle \frac{1}{2021}-1 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-\displaystyle \frac{1}{2021}\quad &&&\textrm{d}.&\displaystyle \frac{2020}{2021}\\\\ \textrm{b}.&-\displaystyle \frac{2020}{2021}&\textrm{c}.&-2021\displaystyle \frac{1}{2}\quad &\textrm{e}.&\displaystyle \frac{1}{2021} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | \displaystyle \frac{1}{2}-1 \right |\times \left | \frac{1}{3}-1 \right |\times \left | \frac{1}{4}-1 \right |\times \left | \frac{1}{5}-1 \right |\times \cdots \times \left | \frac{1}{2021}-1 \right |\\ &=\left | -\displaystyle \frac{1}{2} \right |\times \left | -\frac{2}{3} \right |\times \left | -\frac{3}{4} \right |\times \left | -\frac{4}{5} \right |\times \cdots \times \left | -\frac{2020}{2021} \right |\\ &=\displaystyle \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \cdots \times \frac{2019}{2020}\times \frac{2020}{2021}\\ &=\color{red}\displaystyle \frac{1}{2021} \end{aligned} \end{array}$.

$\begin{array}{ll} 7.&\textrm{Bentuk sederhana dari}\: \: x-5y\: \: \textrm{dan}\\ &5y-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll} \textrm{a}.&\left | 5y-x \right |&\textrm{d}.&\left | y-x \right |\\ \textrm{b}.&\left | y-5x \right |&\textrm{e}.&\left |-5y-x \right |\\ \textrm{c}.&\left | x-5y \right | \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\left | x-5y \right |=m=\begin{cases} \bullet & =x-5y \\ \bullet & =-(x-5y) \end{cases} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: \left | 4m \right |=16\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&\pm 2\\ \textrm{d}.&\pm 4\\ \textrm{e}.&\pm 8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left | 4m \right |&=16\\ (4m)&=\pm 16\\ m&=\pm \displaystyle \frac{16}{4}\\ &=\color{red}\pm 4 \end{aligned}\end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi untuk}\: \: \left | 2x+5 \right |=9\\ & \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&2&&\textrm{d}.&-7\: \: \textrm{dan}\: \: 2\\ \textrm{b}.&2\: \: \textrm{dan}\: \: 7&&\textrm{e}.&-2\: \: \textrm{dan}\: \: 7\\ \textrm{c}.&-7\: \: \textrm{dan}\: \: -2& \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left | 2x+5 \right |&=9\\ \left ( 2x+5 \right )&=\pm 9\\ 2x&=\pm 9-5\\ x&=\displaystyle \frac{\pm 9-5}{2}\\ x&=\color{red}\begin{cases} =\displaystyle \frac{+9-5}{2}=\frac{4}{2}=2 \\\\ =\displaystyle \frac{-9-5}{2}=\frac{-14}{2}=-7 \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: 10-4\left | 4-5x \right |=26\\ & \textrm{adalah}\: ... .\\ &\begin{array}{lllllll}\\ \textrm{a}.&2\: \: \textrm{atau}\: \: 1\displaystyle \frac{2}{3}&&\textrm{d}.&1\: \: \textrm{atau}\: \: -\displaystyle \frac{3}{5}\\\\ \textrm{b}.&2\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}&&\textrm{e}.&-1\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\\ \textrm{c}.&-2\displaystyle \frac{3}{4}\: \: \textrm{atau}\: \: 1\quad&\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}10-4\left | 4-5x \right |&=-26\\ -4\left | 4-5x \right |&=-36\\ \left | 4-5x \right |&=9\\ (4-5x)&=\pm 9\\ -5x&=-4\pm 9\\ x&=\displaystyle \frac{-4\pm 9}{-5}\\ x&=\begin{cases} \displaystyle \frac{-4+9}{-5} & =\color{red}-1 \\ \textrm{atau} & \\ \displaystyle \frac{-4-9}{-5} & = \displaystyle \frac{13}{5}=\color{red}2\frac{3}{5} \end{cases} \end{aligned} \end{array}$

Problem Solving Bentuk Bilangan Riil

Seri Pemecahan Masalah

Jika pada bahasan sebelumnya kita bahas bilangan tidak nyata atau bilangan imajiner pada akar persamaan kuadrat, sekarang kita ketengahkan bahasan sebaliknya, yaitu akar nyta atau riil dari suatu persamaan kuadrat. 

Berikut permasalahannya

(sumber soal dari blog saya sendiri di wordpress)

$\color{blue}\begin{aligned}&\textrm{Akar riil terbesar untuk persamaan}\\ &\color{black}\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^{2}-11x-4\\ &\textrm{adalah}\: \: p+\sqrt{q+\sqrt{r}}\: \:  \textrm{dengan}\: p,\: q,\: \textrm{dan}\: r\: \textrm{adalah}\\ &\textrm{bilangan asli}.\: \: \textrm{Tentukanlah nilai}\: \: p+q+r\\\\ &\color{black}\textbf{Solusi}:\\  \end{aligned}$.

$\begin{aligned}&\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^{2}-11x-4\\ &\frac{3}{x-3}+1+\frac{5}{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1=x^{2}-11x\\ &\frac{3+(x-3)}{x-3}+\frac{5+(x-5)}{x-5}+\frac{17+(x-17)}{x-17}+\frac{19+(x-19)}{x-19}=x^{2}-11x\\ &\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}=x^{2}-11x\\ &\frac{x(x-19)+x(x-3)}{(x-3)(x-19)}+\frac{x(x-17)+x(x-5)}{(x-5)(x-17)}=x^{2}-11x\\ &\frac{2x^{2}-22x}{x^{2}-22x+57}+\frac{2x^{2}-22}{x^{2}-22x+85}=x^{2}-11x\\ &\left ( x^{2}-11x \right )\left ( \frac{2}{x^{2}-22x+57}+\frac{2}{x^{2}-22x+85} \right )\\ &\qquad\qquad\qquad\qquad =x^{2}-11x,\quad \color{red}\textrm{misal}\: \: t=x^{2}-22x\\ &\left ( \frac{2}{t+57}+\frac{2}{t+85} \right )=\frac{x^{2}-11x}{x^{2}-11x}=1\\ &2\left ( t+85 \right )+2\left ( t+57 \right )=(t+57)(t+85)\\ &2t+170+2t+114=t^{2}+142t+4845\\ &0=t^{2}+138t+4731\\ &\color{red}t^{2}+138t+4731=0\: \: \left\{\begin{matrix} a=1\\ b=138\\ c=4731 \end{matrix}\right.\\ &t_{1,2}=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ &t_{1,2}=\displaystyle \frac{-138\pm \sqrt{138^{2}-4.1.4731}}{2}\\ &=\displaystyle \frac{-138\pm \sqrt{19044-18924}}{2}\\ &=\displaystyle \frac{-138\pm \sqrt{120}}{2}\\ &=\displaystyle \frac{-138\pm 2\sqrt{30}}{2}\\ &=-69\pm \sqrt{30} \end{aligned}$.

$\color{red}\begin{aligned}&\color{black}\textrm{Selanjutnya}\\ &t_{1,2}=-69\pm \sqrt{30}\\ &x^{2}-22x=-69\pm \sqrt{30}\\ &x^{2}-22x+69\pm \sqrt{30}=0\\ &x^{2}-22x+69+\sqrt{30}=0\\ &\textrm{atau}\quad x^{2}-22x+69-\sqrt{30}\\ &\\ &\color{black}\textrm{dengan cara yang} \: \: \color{black}\textrm{semisal diatas}\\  &\\ &x_{1,2}=\displaystyle \frac{22\pm \sqrt{22^{2}-4\left ( 69+\sqrt{30} \right )}}{2}\\ &\textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm \sqrt{22^{2}-4\left ( 69-\sqrt{30} \right )}}{2}\\ &x_{1,2}=\displaystyle \frac{22\pm \sqrt{484-276-4\sqrt{30}}}{2}\\ &\textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm \sqrt{484-276+4\sqrt{30}}}{2}\\ &x_{1,2}=\displaystyle \frac{22\pm \sqrt{208-4\sqrt{30}}}{2}\\ &\textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm \sqrt{208+4\sqrt{30}}}{2}\\ &x_{1,2}=\displaystyle \frac{22\pm 2\sqrt{52-\sqrt{30}}}{2}\\ &\textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm 2\sqrt{52+\sqrt{30}}}{2}\\ &x_{1,2}=11\pm \sqrt{52-\sqrt{30}}\\ &\textrm{atau}\qquad x_{3,4}=11\pm \sqrt{52+\sqrt{30}}\\ &\\  &\color{black}\textrm{Maka}, \\ &\left\{\begin{matrix} x_{1}=11+\sqrt{52-\sqrt{30}}\\ \\ x_{2}=11-\sqrt{52-\sqrt{30}} \end{matrix}\right.\\ &\textrm{atau}\qquad \left\{\begin{matrix} x_{3}=11+\sqrt{52+\sqrt{30}}\\ \\ x_{4}=11-\sqrt{52+\sqrt{30}} \end{matrix}\right. \end{aligned}$.

$\begin{aligned}\textrm{Selanjutnya nilai}&\: \textrm{yang paling pas sesuai soal adalah}:\\ &\color{red}x_{3}=11+\sqrt{52+\sqrt{30}}=p+\sqrt{q+\sqrt{r}}\\ \textrm{Sehingga nilai}\: \: \: \: \, \, &p+q+r=11+52+30=93 \end{aligned}$.

Problem Solving Bentuk Bilangan Imajiner (Bilangan Tidak Nyata)

Seri Pemecahan Masalah

Suatu ketika saya sharing-sharing mengenai soal bentuk perpangkatan dari salah seorang teman yang kebetulan memang soalnya membuat penasayaran untuk ditemukan jawabannya.

Berikut soalnya

Saat saya melihat soalnya dengan pangkat berupa angka yang seolah berpola tapi agak susah dicari hungan antara keduanya. Yang satu bilangan utuh yang satu lagi bentuk pecahan (bilangan pada soal, bukan pada yang diketahui). Tapi ada sedikit petunjuk yang mensiratkan soal di atas akan segera dapat dipecahkankan, yaitu posisi yang diketahui  $x+\displaystyle \frac{1}{x}=-1$ adalah salah satu bentuk persamaan kuadrat dengan akar kemungkinan rasional atau imajiner/khayal/tidak nyata dan pangkat pada soal yang semuanya menunjukkan kelipatan 3, yaitu 1234567891011 dan yang satunya posisi penyebut dengan pangkat 1110987654321 dengan basis/bilangan pokok perpangkatannya sama dengan yang diketahui dari soal yaitu  $a$.

Sebelumnya saya pernah menyinggung mengenai istilah definit positif dan definit negatif (silahkan klik di sini) yang kurang lebih istilah tersebut sangat berkaitan dengan akar persamaan kuadrat yang berbentuk imajiner.

Ok, kita kembali ke arah penyelesaian soal di atas, yaitu:

$\begin{aligned}&a+\displaystyle \frac{1}{a}=-1\: \Leftrightarrow\: a^{2}+1=-a\\ &\Leftrightarrow a^{2}+a+1=0\\ &\Leftrightarrow a_{1,2}=\color{red}\displaystyle \frac{-1\pm \sqrt{-3}}{2}=\displaystyle \frac{-1\pm \sqrt{3.(-1)}}{2}\\ &\: \quad\qquad =\displaystyle \frac{-1\pm \sqrt{3}\sqrt{-1}}{2}=\frac{-1\pm \sqrt{3}i}{2}\\ &\: \quad\qquad \: \textrm{dengan}\: \: i=\sqrt{-1}  \end{aligned}$.

$\begin{aligned}&\textrm{Misalkan kita pilih}\: \: a=\color{red}\displaystyle \frac{-1+ \sqrt{3}i}{2}\\ &\textrm{maka nilai dari}\\ &\displaystyle \frac{1}{a}=\displaystyle \frac{1}{\displaystyle \frac{-1+ \sqrt{3}i}{2}}=\displaystyle \frac{2}{-1+ \sqrt{3}i}=\displaystyle \frac{2}{ \sqrt{3}i-1}\\ &\: \quad =\displaystyle \frac{2}{ \sqrt{3}i-1}\times \displaystyle \frac{\sqrt{3}i+1}{\sqrt{3}i+1}=\displaystyle \frac{2(\sqrt{3}i+1)}{-3-1}\\ &\: \quad= -\displaystyle \frac{2(\sqrt{3}i+1)}{-4}=\displaystyle \frac{\sqrt{3}i+1}{-2}\quad \textrm{atau}\\ &\displaystyle \frac{1}{a}=\color{blue}\displaystyle \frac{-1-\sqrt{3}i}{2} \end{aligned}$.

Penjabaran bentuk pangkat dari salah satu akar ternyata membentuk pola yang unik sebagaimana bentuk berikut:

$\begin{aligned}&\begin{cases} a &=\displaystyle \frac{-1+ \sqrt{3}i}{2} \\ \displaystyle \frac{1}{a} & =\displaystyle \frac{-1-\sqrt{3}i}{2} \end{cases},\quad \begin{cases} a^{2} &=\displaystyle \frac{-1- \sqrt{3}i}{2} \\ \displaystyle \frac{1}{a^{2}} & =\displaystyle \frac{-1+\sqrt{3}i}{2} \end{cases}\\ &\qquad\qquad\begin{cases} a^{3} &=1 \\ \displaystyle \frac{1}{a^{3}} & =1 \end{cases}\\ &\begin{cases} a^{4} &=\displaystyle \frac{-1+ \sqrt{3}i}{2} \\ \displaystyle \frac{1}{a^{4}} & =\displaystyle \frac{-1-\sqrt{3}i}{2} \end{cases}\quad \begin{cases} a^{5} &=\displaystyle \frac{-1- \sqrt{3}i}{2} \\ \displaystyle \frac{1}{a^{5}} & =\displaystyle \frac{-1+\sqrt{3}i}{2} \end{cases}\\ &\qquad\qquad\begin{cases} a^{6} &=1 \\ \displaystyle \frac{1}{a^{6}} & =1 \end{cases}\\ &\: \quad\vdots \\ &\cdots \quad \cdots \quad \begin{cases} a^{9} &=1 \\ \displaystyle \frac{1}{a^{9}} & =1 \end{cases}\\ &\cdots \quad \cdots \quad \begin{cases} a^{12} &=1 \\ \displaystyle \frac{1}{a^{12}} & =1 \end{cases}\\ &\cdots \quad \cdots \quad \begin{cases} a^{15} &=1 \\ \displaystyle \frac{1}{a^{15}} & =1 \end{cases}\\ &\textrm{dan seterusnya}\\ & \end{aligned}$.

Jadi, setiap pangkat kelipatan 3 ternyata sama dengan 1, sehingga ini mengakibatkan soal di atas dapat dituliskan lagi dengan

$\begin{aligned}&\color{red}\textrm{Perhatikan lagi bentuk soal}\\ &a^{1234567891011}+\displaystyle \frac{1}{a^{11100987654321}}\\ &=a^{3m}+\displaystyle \frac{1}{a^{3n}}=1+\displaystyle \frac{1}{1}=1+1=\color{red}2 \end{aligned}$.