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

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

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

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

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

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

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

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

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

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

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