Latihan Soal 8 Persiapan PAS Gasal Matematika Wajib Kelas X

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

$\begin{array}{l}\\ 67.& \text{Himpunan penyelesaian dari}\\ & \{\begin{array}{c}x+y+4z=15\\ x-y+z=2\\ x+2y-3z=-4\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{(-1,1,3)\}\\ \text{b}.& \{(1,2,3)\}\\ \text{c}.& \{(-2,1,1)\}\\ \text{d}.& \{(3,2,-1)\}\\ \text{e}.& \{(1,-2,3)\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Semunya dikerjakan dengan metode}\\ & \text{matriks}(\mathbf{\text{Cara Cramer}})\\ & \begin{array}{rl}x& ={\displaystyle \frac{\left|\begin{array}{ccc}15& 1& 4\\ 2& -1& 1\\ -4& 2& -3\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 1\\ 1& 2& -3\end{array}\right|}}\\ & ={\displaystyle \frac{15\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}2& 1\\ -4& -3\end{array}\right|+4\left|\begin{array}{cc}2& -1\\ -4& 2\end{array}\right|}{1\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}}\\ & ={\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}\\ y& ={\displaystyle \frac{\left|\begin{array}{ccc}1& 15& 4\\ 1& 2& 1\\ 1& -4& -3\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 1\\ 1& 2& -3\end{array}\right|}}\\ & ={\displaystyle \frac{1\left|\begin{array}{cc}2& 1\\ -4& -3\end{array}\right|-15\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& 2\\ 1& -4\end{array}\right|}{1\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}}\\ & ={\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}\\ z& ={\displaystyle \frac{\left|\begin{array}{ccc}1& 1& 15\\ 1& -1& 2\\ 1& 2& -4\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 1\\ 1& 2& -3\end{array}\right|}}\\ & ={\displaystyle \frac{1\left|\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right|-1\left|\begin{array}{cc}1& 2\\ 1& -4\end{array}\right|+15\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}{1\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}}\\ & ={\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{array}\end{array}$

$.\text{Cara di atas}$  full matriks-Cramer

$\begin{array}{l}\\ 68.& \text{Hasil dari}xyz\text{yang memenuhi}\\ & \{\begin{array}{c}x+y+z=2\\ x-y+z=-2\\ x-y-z=2\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -8\\ \text{b}.& -4\\ \text{c}.& 2\\ \text{d}.& 4\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=2.....(1)\\ x-y+z=-2.....(2)\\ x-y-z=2.....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =2& \\ x-y+z& =-2& -\\ 2y& =4& \\ y& =2& ....(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =2& \\ x-y-z& =2& +\\ 2x& =4& \\ x& =2& ....(5)\end{array}\\ & \text{Persamaan}(4)\mathrm{\&}(5)\text{ke}(1)\\ & \begin{array}{rl}x+y+z& =2\\ (2)+(2)+z& =2\\ z& =-2\end{array}\\ & \text{Jadi},xyz=(2).(2).(-2)=-8\end{array}\end{array}$

$.\text{Cara di atas}$  full eliminasi-substitusi

$\begin{array}{l}\\ 69.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y+z=-6\\ x-2y+z=3\\ -2x+y+z=9\end{array}\\ & \text{Nilai}xyz=....\\ & \begin{array}{l}\\ \text{a}.& -30\\ \text{b}.& -15\\ \text{c}.& 5\\ \text{d}.& 30\\ \text{e}.& 35\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=-6....(1)\\ x-2y+z=3....(2)\\ -2x+y+z=9....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =-6& \\ x-2y+z& =3& -\\ 3y& =-9& \\ y& =-3& ....(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =-6& \\ -2x+y+z& =9& -\\ 3x& =-15& \\ x& =-5& ....(5)\end{array}\\ & \text{Persamaan}(4)\mathrm{\&}(5)\text{ke}(1)\\ & \begin{array}{rl}x+y+z& =2\\ (-5)+(-3)+z& =-6\\ z& =2\end{array}\\ & \text{Jadi},xyz=(-5).(-3).(2)=30\end{array}\end{array}$

$\begin{array}{l}\\ 70.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+2y+z=4\\ 3x+y+2z=-5\\ x-2y+2z=-6\end{array}\\ & \text{Nilai}xyz=....\\ & \begin{array}{l}\\ \text{a}.& -96\\ \text{b}.& -24\\ \text{c}.& 24\\ \text{d}.& 32\\ \text{e}.& 96\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+2y+z=4.......(1)\\ 3x+y+2z=-5......(2)\\ x-2y+2z=-6.......(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+2y+z& =4& |\times 1|& x+2y+z& =4\\ 3x+y+2z& =-5& |\times 2|& 6x+2y+4z& =-10& -\\ & & & -5x-3z& =14& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+2y+z& =4& \\ x-2y+2z& =-6& +\\ 2x+3z& =-2& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ -5x-3z& =14& \\ 2x+3z& =-2& +\\ -3x& =12& \\ x& =-4& .....(6)\\ \text{didapat pula}& z& =2......(7)\end{array}\\ & \text{Dari persamaan}(6)\mathrm{\&}(7)\text{didapatkan}\\ & \begin{array}{rl}x+2y+z& =4\\ (-4)+2y+2& =4\\ y& =3\end{array}\\ & \text{Jadi},xyz=(-4).(3).(2)=-24\end{array}\end{array}$.

$\begin{array}{l}\\ 71.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y-z=1\\ 2x-y+2z=9\\ x+3y-z=7\end{array}\\ & \text{Nilai}{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}}\\ \text{b}.& {\displaystyle \frac{3}{4}}\\ \text{c}.& {\displaystyle \frac{13}{12}}\\ \text{d}.& {\displaystyle \frac{5}{4}}\\ \text{e}.& {\displaystyle \frac{7}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y-z=1....(1)\\ 2x-y+2z=9....(2)\\ x+3y-z=7....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y-z& =1& \\ 2x-y+2z& =9& +\\ 3x+z& =10& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y-z& =1& |\times 3|& 3x+3y-3z& =3& \\ x+3y-z& =7& |\times 1|& x+3y-z& =7& -\\ & & & 2x-2z& =-4& \\ & & & x-z& =2& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ 3x+z& =10& \\ x-z& =-2& +\\ 4x& =8& \\ x& =2& .....(6)\\ \text{didapat pula}& z& =4......(7)\end{array}\\ & \text{Dari persamaan}(1)\mathrm{\&}(3)\text{didapatkan juga}\\ & \begin{array}{l}\\ x+y-z& =1& \\ x+3y-z& =-7& -\\ -2y& =-6& \\ y& =3& ....(8)\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}={\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}}}\end{array}\end{array}$

$\begin{array}{l}\\ 72.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y+z=5\\ x+y-4z=10\\ -2x+y+z=0\end{array}\\ & \text{Nilai dari}{\displaystyle \frac{xz}{y}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\frac{6}{13}}\\ \text{b}.& {\displaystyle -\frac{5}{13}}\\ \text{c}.& {\displaystyle -\frac{1}{13}}\\ \text{d}.& {\displaystyle \frac{1}{13}}\\ \text{e}.& {\displaystyle \frac{7}{13}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=5.....(1)\\ x+y-4z=10.....(2)\\ -2x+y+z=0.....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =5& \\ x+y-4z& =10& -\\ 5z& =-5& \\ z& =-1& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =5& \\ -2x+y+z& =0& -\\ 3x& =5& \\ x& ={\displaystyle \frac{5}{3}}& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{rl}x+y+z& =5\\ {\displaystyle \frac{5}{3}+y-1}& =5\\ y& =5+1-{\displaystyle \frac{5}{3}=\frac{13}{3}}\end{array}\\ & \text{Jadi},{\displaystyle \frac{xz}{y}={\displaystyle \frac{\left({\displaystyle \frac{5}{3}}\right).(-1)}{{\displaystyle \frac{13}{3}}}=-{\displaystyle \frac{5}{13}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 73.& \text{Himpunan penyelesaian dari}\\ & \{\begin{array}{ll}{\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4\end{array}\\ & \text{adalah}\{(x,y,z)\},\text{maka}x+3z=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& {\displaystyle \frac{1}{3}}\\ \text{c}.& 1\\ \text{d}.& 3\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{ll}{\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8....(1)\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10.....(2)\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4...........(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10& -\\ -{\displaystyle \frac{1}{x}-\frac{1}{z}}& =-2\\ {\displaystyle \frac{1}{x}+\frac{1}{z}}& =2& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =8& |\times 2|& {\displaystyle \frac{4}{x}+\frac{4}{y}+\frac{8}{z}}& =16\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4& |\times 1|& {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4& -\\ & & & {\displaystyle \frac{2}{x}+\frac{6}{z}}& =12\\ & & \iff & {\displaystyle \frac{1}{x}+\frac{3}{z}}& =6& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{x}+\frac{3}{z}}& =6\\ {\displaystyle \frac{1}{x}+\frac{1}{z}}& =2-\\ {\displaystyle \frac{2}{z}}& =4& \\ z& ={\displaystyle \frac{1}{2}......(6)}\\ x& =2-{\displaystyle \frac{1}{2}={\displaystyle \frac{3}{2}}}\end{array}\\ & \text{Jadi},x+3z={\displaystyle \frac{3}{2}+3.\frac{1}{2}=3}\end{array}\end{array}$.

$\begin{array}{l}\\ 74.& \text{Diketahui tiga buah bilangan berturut-turut}\\ & a,b,\text{dan}c.\text{Rata-rata dari ke tiga bilangan}\\ & \text{itu adalah 12. Bilangan kedua sama dengan}\\ & \text{jumlah bilangan yang lain dikurangi 12}.\\ & \text{Jika bilangan ke tiga sama dengan jumlah}\\ & \text{bilangan yang lain, maka nilai}2a+b-c=....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle 42}\\ \text{b}.& -{\displaystyle 36}\\ \text{c}.& -{\displaystyle 18}\\ \text{d}.& -{\displaystyle 12}\\ \text{e}.& -{\displaystyle 6}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Model matematika dari persamaan di atas}\\ & \{\begin{array}{c}a+b+c=36....(1)\\ -a+b-x=12....(2)\\ a+b-c=0....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ a+b+c& =36& \\ -a+b-c& =12& +\\ 2b& =48& \\ b& =24& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ a+b+c& =36& \\ a+b-c& =0& -\\ 2c& =36& \\ c& =18& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{rl}a+b+c& =36\\ a+24+18& =36\\ a& =36-42\\ & =-6\end{array}\\ & \text{Jadi},2a+b-c=2(-6)+24-18=-6\end{array}\end{array}$

$\begin{array}{l}\\ 75.& \text{Jumlah uang terdiri atas koin pecahan}Rp500,00\\ & Rp200,00danRp100,00\text{dengan nilai total}\\ & Rp100.000,00.\text{Jika nilai uang pecahan 500-an}\\ & \text{setengah dari nilai uang pecahan 200-an, tetapi}\\ & \text{tiga kali uang pecahan 100-an, maka banyak koin}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 460\\ \text{b}.& 440\\ \text{c}.& 420\\ \text{d}.& 380\\ \text{e}.& 350\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Model matematika dari kasus di atas}\\ & \{\begin{array}{c}A(500)+B(200)+C(100)=100.000....(1)\\ A(500)={\displaystyle \frac{1}{2}B(200)....(2)}\\ A(500)=3C(100)....(3)\end{array}\\ & \text{Dari persamaan}(2)\text{didapatkan}\\ & 2A(500)=B(200)\\ & \text{Dari persamaan}(3)\text{akan didapatkan}\\ & {\displaystyle \frac{1}{3}A(500)=C(100)}\\ & \text{Dari persamaan}(1)\text{maka},\\ & A(500)+B(200)+C(100)=100.000\\ & A(500)+2A(500)+{\displaystyle \frac{1}{3}A(500)=100.000}\\ & {\displaystyle \frac{10}{3}A(500)=100.000\iff A(500)=30.000}\\ & \text{maka akan didapatkan}\\ & B(200)=2(30.000)=60.000\\ & C(100)={\displaystyle \frac{1}{3}(30.000)=10.000}\\ & \{\begin{array}{ll}A(500)& =30.000\Rightarrow A={\displaystyle \frac{30.000}{500}=60}\\ B(200)& =60.000\Rightarrow B={\displaystyle \frac{60.000}{200}=300}\\ C(100)& =10.000\Rightarrow C={\displaystyle \frac{10.000}{100}=100}\end{array}\\ & \text{Jadi},A+B+C=60+300+100=460\end{array}\end{array}$.

Latihan Soal 7 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{l}\\ 56.& \text{Nilai}x\text{berikut yang tidak memenuhi}\\ & {\displaystyle \frac{x-3}{x^2+2x+1}\le 0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -1\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {\displaystyle \frac{x-3}{x^2+2x+1}\le 0}\\ & {\displaystyle \frac{(x-3)}{(x+1{)}^2}\le 0}\\ & \text{Pembuat nol}\\ & \{\begin{array}{l}x=3,\text{boleh digunakan}\\ x=-1,\text{tetapi}x\ne -1,\end{array}\\ & \text{sehingga}-1\text{tidak digunakan}\\ & \begin{array}{c}\\ & & & & & & & & & & \\ & -& -& -& -& -& & +& +& & \\ & & -1& & & & 3& & & & \\ & & & & & & & & & & \end{array}& \end{array}\end{array}$.

$\begin{array}{l}\\ 57.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x-5}\le x\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 1<x<0\\ \\ \text{b}.& x<1\text{atau}x\ge 5\\ \\ \text{c}.& {\displaystyle \frac{5}{6}\le x\le 1\text{atau}x\ge 5}\\ \\ \text{d}.& {\displaystyle \frac{5}{6}\le x<1\text{atau}5<x<6}\\ \\ \text{e}.& x\ge 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{6x-5}& \le x\\ 1.\text{Kuadrat}& \text{kan}\\ 6x-5& \le x^2\\ -x^2+6x-5& \le 0\\ x^2-6x+5& \ge 0\\ (x-1)(x-5)& \ge 0\\ x\le 1\text{atau}& x\ge 5\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x-5& \ge 0\\ 6x& \ge 5\\ x& \ge {\displaystyle \frac{5}{6}}\end{array}\end{array}$

$\begin{array}{l}\\ 58.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x+6}>6\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x>7\\ \text{b}.& x\ge 7\\ \text{c}.& x<7\\ \text{d}.& x>1\\ \text{e}.& x\ge 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\sqrt{6x+6}& >6\\ 1.\text{Kuadrat}& \text{kan}\\ 6x+6& >36\\ x+1& >6\\ x& >7\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x+6& \ge 0\\ 6x& \ge 6\\ x& \ge {\displaystyle \frac{6}{6}}\\ x& \ge 1\end{array}\end{array}$

$\begin{array}{l}\\ 59.& \text{Penyelesaian pertidaksamaan}\\ & x+2>{\displaystyle \sqrt{10-x^2}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\le x\le \sqrt{10}\\ \text{b}.& 1<x\le \sqrt{10}\\ \text{c}.& -3<x\le \sqrt{10}\\ \text{d}.& -\sqrt{10}\le x\le \sqrt{10}\\ \text{e}.& x<-3\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}x+2& >{\displaystyle \sqrt{10-x^2}}\\ 1.\text{Kuadrat}& \text{kan}\\ x^2+4x+4& >10-x^2\\ 2x^2+4x& +4-10>0\\ 2x^2+4x-& 6>0\\ x^2+2x-& 3>0\\ (x+3)& (x-1)>0\\ x<-3& \text{atau}x>1\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 10-x^2& \ge 0\\ x^2-10& \le 0\\ (x-\sqrt{10})& (x+\sqrt{10})\le 0\\ -\sqrt{10}\le x& \le \sqrt{10}\end{array}\end{array}$

$\begin{array}{l}\\ 60.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{3x+7}\ge x-1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -1<x<6\\ \text{b}.& -1\le x<6\\ \text{c}.& x\ge -{\displaystyle \frac{7}{3}}\\ \text{d}.& -{\displaystyle \frac{7}{3}\le x\le 6}\\ \text{e}.& -{\displaystyle \frac{7}{3}\le x\le 1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{3x+7}& \ge x-1\\ 1.\text{Kuadrat}& \text{kan}\\ 3x+7& \ge x^2-2x+1\\ -x^2+3x& +2x+7-1\ge 0\\ -x^2+5x+& 6\ge 0\\ x^2-5x-& 6\le 0\\ (x+1)& (x-6)\le 0\\ -1\le x& \le 6\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 3x+7& \ge 0\\ 3x& \ge -7\\ x& \ge -{\displaystyle \frac{7}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 61.& \text{Nilai}x\text{yang memenuhi}\\ & \sqrt{2x-8}<\sqrt{x+5}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x\ge -5\\ \text{b}.& x<-13\text{atau}x\ge -4\\ \text{c}.& x<13\\ \text{d}.& 4\le x<13\\ \text{e}.& -5\le x\le 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{2x-8}& <\sqrt{x+5}\\ (1)\text{kuadratkan}& \\ 2x-8& <x+5\\ x& <13\\ (2)2x-8\ge 0& \\ x& \ge 4\\ (3)x+5\ge 0& \\ x& \ge -5\\ \text{perhatikan}& \text{lah garis bilangannya berikut}\end{array}\end{array}$.

$\begin{array}{l}\\ 62.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x-4}<\sqrt{2x+8}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -4<x\le {\displaystyle \frac{2}{3}}\\ \text{b}.& -4<x<3\\ \text{c}.& {\displaystyle \frac{2}{3}\le x<3}\\ \text{d}.& 2<x\le 4\\ \text{e}.& -4\le x\le 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \sqrt{6x-4}}& <\sqrt{2x+8}\\ 1.\text{Kuadrat}& \text{kan}\\ 6x-4& <2x+8\\ 6x-2x& <8+4\\ 4x& <12\\ x& <3\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x-4& \ge 0\\ 6x& \ge 4\\ x& \ge {\displaystyle \frac{2}{3}}\\ 3.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 2x+8& \ge 0\\ 2x& \ge -8\\ x& \ge -4\end{array}\end{array}$

$\begin{array}{l}\\ 63.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{x+3}>\sqrt{12-2x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 3<x\le 6\\ \text{b}.& -3<x\le 6\\ \text{c}.& -6<x\le 3\\ \text{d}.& -6<x\le -3\\ \text{e}.& x<3\text{atau}x>6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle {\displaystyle \sqrt{x+3}}}& >\sqrt{12-2x}\\ 1.\text{Kuadrat}& \text{kan}\\ x+3& >12-2x\\ x+2x& >12-3\\ 3x& >9\\ x& >3\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ x+3& \ge 0\\ x& \ge -3\\ 3.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 12-2x& \ge 0\\ 2x-12& \le 0\\ 2x& \le 12\\ x& \le 6\end{array}\end{array}$

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

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

Latihan Soal 6 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 46.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{2}{x+1}\le \left|x\right|\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& \{x|x\le -2\text{atau}x\ge 1\}\\ \text{b}.& \{x|x\le -2\text{atau}0\le x\le 1\}\\ \text{c}.& \{x|x\ge 1\}\\ \text{d}.& \{x|x<-1\text{atau}x\ge 1\}\\ \text{e}.& \{x|-1<x\le 1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \left|x\right|\ge {\displaystyle \frac{2}{x+1}\text{berakibat}}\\ & {\displaystyle \frac{-2}{x+1}\ge x\text{atau}x\ge {\displaystyle \frac{2}{x+1}}}\\ & \bullet \text{bagian}1\\ & x\le {\displaystyle \frac{-2}{x+1}(\mathbf{\text{tidak boleh kali silang}})}\\ & x+{\displaystyle \frac{2}{x+1}\le 0}\\ & {\displaystyle \frac{x(x+1)+2}{x+1}\le 0}\\ & {\displaystyle \frac{x^2+x+2}{x+1}\le 0\iff {\displaystyle \frac{\text{Definit positif}}{x+1}\le 0}}\\ & \begin{array}{rl}& {\text{HP}}_1=\{x|x<-1,x\in \mathbb{R}\}\\ & \bullet \text{bagian}2\\ & x\ge {\displaystyle \frac{2}{x+3}}\\ & x-{\displaystyle \frac{2}{x+1}\ge 0}\\ & {\displaystyle \frac{x(x+1)-2}{x+1}\ge 0}\\ & {\displaystyle \frac{x^2+x-2}{x+1}\ge 0}\\ & {\displaystyle \frac{(x+2)(x-1)}{x+1}\ge 0}\\ & {\text{HP}}_2=\{x|-2\le x<-1\text{atau}x\ge 1,x\in \mathbb{R}\}\\ & \text{HP}={\text{HP}}_1+{\text{HP}}_2=\{x|x<-1\text{atau}x\ge 1,x\in \mathbb{R}\}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 47.& \text{Diketahui pertidaksamaan}{\displaystyle \frac{x+10}{x-9}\le 0}\\ & \text{dan diberikan beberapa nilai berikut}\\ & (\text{i})x=-6(\text{iii})x=-14\\ & (\text{ii})x=-10(\text{iv})x=-18\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \text{di atas adalah ditunjukkan oleh}....\\ & \begin{array}{l}\\ \text{a}.& (\text{i})\text{dan}(\text{ii})\\ \text{b}.& (\text{i})\text{dan}(\text{iii})\\ \text{c}.& (\text{ii})\text{dan}(\text{iii})\\ \text{d}.& (\text{ii})\text{dan}(\text{iv})\\ \text{e}.& (\text{iii})\text{dan}(\text{iv})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{x+10}{x-9}}& \le 0\\ \text{HP}=& \{x|-10\le x<9,x\in \mathbb{R}\}\end{array}\end{array}$.

$\begin{array}{l}\\ 48.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\text{ adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 2\le x<6\\ \text{b}.& 2\le x<3\\ \text{c}.& 2<x<3\text{atau}x>6\\ \text{d}.& x<3\text{atau}3<x<6\\ \text{e}.& x<2\text{atau}x>3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {\displaystyle \frac{6}{x-3}<\frac{8}{x-2}}\\ & \iff {\displaystyle \frac{6}{x-3}-\frac{8}{x-2}<0}\\ & \iff {\displaystyle \frac{6(x-2)-8(x-3)}{(x-3)(x-2)}<0}\\ & \iff {\displaystyle \frac{6x-8x-12+24}{(x-2)(x-3)}<0}\\ & \iff {\displaystyle \frac{-2x+12}{(x-2)(x-3)}<0}\\ & \iff {\displaystyle \frac{2x-12}{(x-2)(x-3)}>0}\\ & \iff {\displaystyle \frac{2(x-6)}{(x-2)(x-3)}>0}\\ & \text{HP}=\{x|2<x<3\text{atau}x>6,x\in \mathbb{R}\}\end{array}\end{array}$.

$\begin{array}{l}\\ 49.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \frac{x^2-81}{x^2}\ge 0\text{ adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x\le -9\text{atau}x\ge 9\\ \text{b}.& -9\le x<0\text{atau}x\ge 9\\ \text{c}.& -9\le x<0\text{atau}0<x\le 9\\ \text{d}.& -9<x\le 9\\ \text{e}.& x\in \mathbb{R}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{x^2-81}{x^2}\ge 0}\\ & {\displaystyle \frac{(x+9)(x-9)}{x^2}\ge 0}\\ & \text{HP}=\{x|x\le -9\text{atau}x\ge 9,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 50.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2-4}{x+2}>0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x>2\\ \text{b}.& -2\le x<2\\ \text{c}.& x<-2\text{atau}x>2\\ \text{d}.& x<-2\text{atau}-2<x<2\\ \text{e}.& x\ge -2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{x^2-4}{x+2}>0}\\ & {\displaystyle \frac{(x+2)(x-2)}{(x+2)}>0}\\ & (x-2)>0\\ & x>2\end{array}\end{array}$.

$\begin{array}{l}\\ 51.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2+x-30}{2x^2+13x-45}<0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& \{x|-9<x<5,x\in \mathbb{R}\}\\ \text{b}.& \{x|-6<x<5,x\in \mathbb{R}\}\\ \text{c}.& \{x|-9<x<-6\text{atau}x<5,x\in \mathbb{R}\}\\ \text{d}.& \{x|-9<x<-6\text{atau}{\displaystyle \frac{5}{2}<x<5,x\in \mathbb{R}}\}\\ \text{e}.& \{x|x<-9\text{atau}-6<x<{\displaystyle \frac{5}{2}\text{atau}x<5,x\in \mathbb{R}}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2+x-30}{2x^2+13x-45}}& <0\\ {\displaystyle \frac{(x+6)(x-5)}{(x+9)(2x-5)}}& <0\\ \text{Cukup jelas}& \end{array}\end{array}$.

$\begin{array}{l}\\ 52.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{2x+6}{x-4}\le 1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -10<x<4\\ \text{b}.& -10\le x<4\\ \text{c}.& -4<x\le 10\\ \text{d}.& x\le -10\text{atau}x\ge 4\\ \text{e}.& x<-10\text{atau}x\ge 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{2x+6}{x-4}}& \le 1\\ {\displaystyle \frac{2x+6}{x-4}-1}& \le 0\\ {\displaystyle \frac{2x+6-(x-4)}{x-4}}& \le 0\\ {\displaystyle \frac{x+10}{x-4}}& \le 0\end{array}\end{array}$

$\begin{array}{l}\\ 53.& \text{Nilai}x\text{yang memenuhi}{\displaystyle \frac{x^2-x}{x+3}\ge 1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x<-3\text{atau}-1\le x\le 3\\ \text{b}.& -3<x\le -1\text{atau}x\ge 3\\ \text{c}.& -3\le x\le 3\\ \text{d}.& -3\le x\le -1\text{atau}x\ge 3\\ \text{e}.& -3\le x\le -1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2-x}{x+3}}& \ge 1\\ {\displaystyle \frac{x^2-x}{x+3}}& -1\ge 0\\ {\displaystyle \frac{x^2-x-(x+3)}{x+3}}& \ge 0\\ {\displaystyle \frac{x^2-2x-3}{x+3}}& \ge 0\\ {\displaystyle \frac{(x-3)(x+1)}{x+3}}& \ge 0\end{array}\end{array}$

$\begin{array}{l}\\ 54.& \text{Nilai}x\text{yang memenuhi}\\ & x+2+{\displaystyle \frac{1}{x+4}>0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x<-4\text{atau}x\ge -3\\ \text{b}.& x<-4\text{atau}x>-3\\ \text{c}.& -4\le x\le -3\\ \text{d}.& x>-4\\ \text{e}.& -4\le x\le -3\text{atau}x>-3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}x+2+{\displaystyle \frac{1}{x+4}}& >0\\ {\displaystyle \frac{(x+2)(x+4)+1}{(x+4)}}& >0\\ {\displaystyle \frac{x^2+6x+8+1}{x+4}}& >0\\ {\displaystyle \frac{x^2+6x+9}{x+4}}& >0\\ {\displaystyle \frac{(x+3{)}^2}{(x+4)}}& >0\\ x& >-4\end{array}\end{array}$

$\begin{array}{l}\\ 55.& \text{Nilai}x\text{yang memenuhi}\\ & x+3<{\displaystyle \frac{x^2+6x+11}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-3{\displaystyle \frac{2}{3}\text{atau}x>0,x\in \mathbb{R}}\}\\ \text{b}.& \{x|0\le x\le 11,x\in \mathbb{R}\}\\ \text{c}.& \{x|x<-11\text{atau}x>0,x\in \mathbb{R}\}\\ \text{d}.& \{x|x<0\text{atau}x>11,x\in \mathbb{R}\}\\ \text{e}.& \{x|x\le 11\text{atau}x>0,x\in \mathbb{R}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}x+3<{\displaystyle \frac{x^2+6x+11}{x}}& \\ x+3-{\displaystyle \frac{x^2+6x+11}{x}}& <0\\ {\displaystyle \frac{x(x+3)-(x^2+6x+11)}{x}}& <0\\ {\displaystyle \frac{x^2+3x-x^2-6x-11}{x}}& <0\\ {\displaystyle \frac{-3x-11}{x}}& <0\\ {\displaystyle \frac{3x+11}{x}}& >0\end{array}\end{array}$.

Latihan Soal 5 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{l}\\ 36.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |3-\left|x\right||<10\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-14\text{atau}x>12\\ \text{b}.& x<-13\text{atau}x>13\\ \text{c}.& x<-12\text{atau}x>10\\ \text{d}.& 0<x<10\\ \text{e}.& -13<0<13\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& |3-\left|x\right||<10\\ & -10<3-\left|x\right|<10\\ & -13<-\left|x\right|<7\\ & -7<\left|x\right|\le 13,(\text{ingat harga}\left|x\right|\ge 0)\\ & 0\le \left|x\right|<13\\ & \text{selanjutnya},\\ & \left|x\right|<13\\ & -13<x<13\\ & \text{HP}=\{x|-13<x<13,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \text{(UM UGM 05)}\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |x^2-3|<2x\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -1<x<3\\ \text{b}.& -3<x<1\\ \text{c}.& 1<x<3\\ \text{d}.& -3<x<-1\text{atau}1<x<3\\ \text{e}.& x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}|x^2-3|& <2x\\ -2x<(x^2-3)& <2x\\ \text{dipartisi men}& \text{jadi dua bagian}\\ \bullet \text{pertama}& \\ (x^2-3)& >-2x\\ x^2+2x-3& >0\\ (x+3)(x-1)& >0\\ x<-3\text{atau}& x>1\\ \bullet \text{kedua}& \\ (x^2-3)& <2x\\ x^2-2x-3& <0\\ (x-3)(x+1)& <0\\ -1<x<3& \\ \text{ambil yang}& \text{memenuhi keduanya}\\ \text{berupa iris}& \text{an}\\ \text{HP}=& \{1<x<3,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 38.& (\text{SPMB 05})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {|x-2|}^2<4|x-2|+12\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \{x\in \mathbb{R}|2\le x\le 8\}\\ \text{b}.& \{x\in \mathbb{R}|4<x<8\}\\ \text{c}.& \{x\in \mathbb{R}|-4<x<8\}\\ \text{d}.& \{x\in \mathbb{R}|-2<x<4\}\\ \text{e}.& \{x\in \mathbb{R}|2<x<4\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{misalkan}p& =|x-2|,\text{selanjutnya}\\ {|x-2|}^2<& 4|x-2|+12\\ p^2<& 4p+12\\ p^2-& 4p-12<0\\ (p-6)& (p+2)<0\\ -2<p& <6,\text{atau jika dikembalikan}\\ -2<& |x-2|<6,\\ & \text{ingat, nilanya tidak negatif}\\ 0\le & |x-2|<6\\ -6<& x-2<6\\ -4<& x<8\\ \text{HP}=& \{-4<x<8,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Diketahui grafik fungsi}f(x)=m{x}^2-2mx+m\\ & \text{berada di atas grafik fungsi}\\ & g(x)=2x^2-3,\text{maka nilai}m\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& m>2& & & \text{d}.& -6<m<2\\ \text{b}.& m>6& \text{c}.& 2<m<6& \text{e}.& m<-6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(x)& =g(x)\\ m{x}^2-2mx+m& =2x^2-3\\ m{x}^2-2x^2-2mx+m+3& =0\\ \text{Supaya grafik}f(x)\text{berada }& \text{di atasnya},\\ \text{maka}D=B^2-4AC& <0\\ (m-2)x^2-2mx+(m+3)& =0\{\begin{array}{ll}A& =m-2\\ B& =-2m\\ C& =m+3\end{array}\\ B^2-4AC& <0\\ (-2m{)}^2-4(m-2)(m+3)& <0\\ 4m^2-4(m^2+m-6)& <0\\ 4m^2-4m^2-4m+24& <0\\ -4m+24& <0\\ m-6& >0\\ m& >6\end{array}\end{array}$.

$\begin{array}{l}\\ 40.& \text{Jika}3<x<5\text{maka penyelesaian untuk}\\ & \sqrt{x^2-6x+9}-\sqrt{x^2-10x+25}=...\\ & \begin{array}{l}\\ \text{a}.& 2x-2& & & \text{d}.& -2\\ \\ \text{b}.& 2& \text{c}.& 8-2x& \text{e}.& 2x-8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \sqrt{x^2-6x+9}-\sqrt{x^2-10x+25}\\ & =\sqrt{(x-3{)}^2}-\sqrt{(x-5{)}^2}\\ & =|x-3|-|x-5|\\ & \text{ingat bahwa saat}3<x<5\\ & \text{maka}\{\begin{array}{l}|x-3|=(x-3)\\ |x-5|=-(x-5)\end{array},\\ & \text{sehingga}\\ & =|x-3|-|x-5|=(x-3)-(-(x-5))\\ & =x-3+x-5\\ & =2x-8\end{array}\end{array}$.

$\begin{array}{l}\\ 41.& \text{Jika}1<x<5\text{maka penyelesaian untuk}\\ & \sqrt{x^2-2x+1}+\sqrt{x^2-10x+25}=...\\ & \begin{array}{l}\\ \text{a}.& 2& & & \text{d}.& 5\\ \\ \text{b}.& 3& \text{c}.& 4& \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \sqrt{x^2-2x+1}+\sqrt{x^2-10x+25}\\ & =\sqrt{(x-1{)}^2}+\sqrt{(x-5{)}^2}\\ & =|x-1|+|x-5|\\ & \text{ingat bahwa saat}1<x<5\\ & \text{maka}\{\begin{array}{l}|x-1|=(x-1)\\ |x-5|=-(x-5)\end{array},\\ & \text{sehingga}\\ & =|x-1|+|x-5|=(x-1)+(-(x-5))\\ & =x-1+5-x\\ & =4\end{array}\end{array}$.

$\begin{array}{l}\\ 42.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{5}{4x-3}}\right|\le 1\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\le x\le \frac{3}{4}\text{atau}x\ge 2}\\ \text{b}.& x\le -{\displaystyle \frac{1}{2}\text{atau}{\displaystyle \frac{3}{4}<x\le 2}}\\ \text{c}.& -{\displaystyle \frac{1}{2}\le x\le 2,x\ne \frac{3}{4}}\\ \text{d}.& x\le -{\displaystyle \frac{1}{2}\text{atau}x>\frac{3}{4}}\\ \text{e}.& x\le -{\displaystyle \frac{1}{2}\text{atau}x\ge 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{5}{4x-3}}\right|& \le 1\\ -1\le {\displaystyle \frac{5}{4x-3}}& \le 1,\mathbf{\text{jika dibalik}}\\ -1\ge {\displaystyle \frac{4x-3}{5}}& \ge 1,\mathbf{\text{bentuk ini tidak}}\\ \mathbf{\text{dibolehkan}}& \mathbf{\text{maka perlu diubah menjadi}}\\ -1\ge {\displaystyle \frac{4x-3}{5}\text{atau}}& {\displaystyle \frac{4x-3}{5}\ge 1,\text{selanjutnya}}\\ \bullet \text{bagian}& 1\\ -1& \ge {\displaystyle \frac{4x-3}{5}\iff \frac{4x-3}{5}\le -1}\\ 4x-3& \le -5\\ 4x& \le -2\\ x& \le -{\displaystyle \frac{1}{2}}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{4x-3}{5}}& \ge 1\\ 4x-3& \ge 5\\ 4x& \ge 8\\ x& \ge 2\end{array}\end{array}$

$\begin{array}{l}\\ 43.& (\text{UMPTN 95})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{2}{2x-1}}\right|>1\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x>2\\ \text{b}.& x<2\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{c}.& x<-1\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{d}.& -1<x<2\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{e}.& x<-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{semua opsi bukan jawaban}}\\ & \mathbf{\text{Berikut pembahasannya}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{2}{2x-1}}\right|& >1\\ -1>{\displaystyle \frac{2}{2x-1}}& \text{atau}{\displaystyle \frac{2}{2x-1}>1,\mathbf{\text{dibalik}}}\\ -1<{\displaystyle \frac{2x-1}{2}}& \text{atau}{\displaystyle \frac{2x-1}{2}<1}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{2x-1}{2}}& >-1\\ 2x-1& >-2\\ 2x& >-1\\ x& >-{\displaystyle \frac{1}{2}}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{2x-1}{2}}& <1\\ 2x-1& <2\\ 2x& <3\\ x& <{\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 44.& (\text{UMPTN 00})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{2x+7}{x-1}}\right|\ge 1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2\le x\le 8\\ \text{b}.& x\le -8\text{atau}x\ge -2\\ \text{c}.& -8\le x<1\text{atau}x>1\\ \text{d}.& -2\le x<1\text{atau}1<x\le 8\\ \text{e}.& x\le -8\text{atau}-2\le x<1\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{2x+7}{x-1}}\right|& \ge 1\\ -1\ge {\displaystyle \frac{2x+7}{x-1}}& \text{atau}{\displaystyle \frac{2x+7}{x-1}\ge 1}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{2x+7}{x-1}}& \le -1(\mathbf{\text{tidak boleh kali silang}})\\ {\displaystyle \frac{2x+7}{x-1}}& +1\le 0\\ & {\displaystyle \frac{2x+7+(x-1)}{x-1}\le 0}\\ {\displaystyle \frac{3x+6}{x-1}}& \le 0\\ {\text{HP}}_1=& \{x|-2\le x<1,x\in \mathbb{R}\}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{2x+7}{x-1}}& \ge 1\\ {\displaystyle \frac{2x+7}{x-1}}& -1\ge 0\\ & {\displaystyle \frac{2x+7-(x-1)}{x-1}\ge 0}\\ {\displaystyle \frac{x+8}{x-1}}& \ge 0\\ {\text{HP}}_2=& \{x|x\le -8\text{atau}x>1,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x\le -8\text{atau}-2\le x<1\text{atau}x>1,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 45.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{x-2}{x+3}}\right|\le 2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -8\le x<-3\\ \text{b}.& -8\le x<-1\\ \text{c}.& -4\le x<-3\\ \text{d}.& x\le -8\text{atau}x\ge -{\displaystyle \frac{4}{3}}\\ \text{e}.& x\le -4\text{atau}x\ge 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{x-2}{x+3}}\right|& \le 2\\ -2\le {\displaystyle \frac{x-2}{x+3}}& \text{atau}{\displaystyle \frac{x+2}{x+3}\le 2}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{x-2}{x+3}}& \ge -2(\mathbf{\text{tidak boleh kali silang}})\\ {\displaystyle \frac{x-2}{x+3}}& +2\ge 0\\ & {\displaystyle \frac{x-2+2(x+3)}{x+3}\ge 0}\\ {\displaystyle \frac{3x+4}{x+3}}& \ge 0\\ {\text{HP}}_1=& \{x|x<-3\text{atau}x\ge -{\displaystyle \frac{4}{3},x\in \mathbb{R}}\}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{x-2}{x+3}}& \le 2\\ {\displaystyle \frac{x-2}{x+3}}& -2\le 0\\ & {\displaystyle \frac{x-2-2(x+3)}{x+3}\le 0}\\ {\displaystyle \frac{-x-8}{x+3}}& \le 0,\mathbf{\text{koefisien textit{x} negatif}}\\ {\displaystyle \frac{x+8}{x+3}}& \ge 0\\ {\text{HP}}_2=& \{x|x\le -8\text{atau}x>-3,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x\le -8\text{atau}x\ge -{\displaystyle \frac{4}{3},x\in \mathbb{R}}\}\end{array}\end{array}$

Latihan Soal 4 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 26.& \text{Nilai}x\text{yang memenuhi}\\ & (x+3)(x-1)\ge (x-1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\le x\le 3& \\ \text{b}.& x\le -2\text{atau}x\ge 1\\ \text{c}.& -3\le x\le -1\\ \text{d}.& -2\ge x\text{atau}x\ge 3\\ \text{e}.& -1\ge x\text{atau}x\ge 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}(x+3)(x-1)& \ge (x-1)\\ (x+3)(x-1)-(x-1)& \ge 0\\ (x-1)((x+3)-1)& \ge 0\\ (x-1)(x+2)& \ge 0\end{array}\\ & \text{Sehingga solusinya adalah:}\\ & x\le -2\text{atau}x\ge 1\end{array}$.

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

$\begin{array}{l}\\ 28.& \text{Himpunan penyelesaian pertidaksamaan}\\ & |x^2+5x|\le 6\text{adalah... .}\\ & \text{a}.\{x|-6\le x\le 1\}\\ & \text{b}.\{x|-3\le x\le -2\}\\ & \text{c}.\{x|-6\le x\le -3\text{atau}-2\le x\le 1\}\\ & \text{d}.\{x|-6\le x\le -5\text{atau}-2\le x\le 0\}\\ & \text{e}.\{x|-5\le x\le -3\text{atau}-2\le x\le 0\}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{r}\text{Diketahui bahwa}\\ & \\ |x^2+5x|& \le 6\\ -6\le x^2+5x& \le 6\\ & \end{array}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}& -6\le x^2+5x\\ & x^2+5x+6\ge 0\\ & (x+3)(x+2)\ge 0\end{array}& \begin{array}{rl}& x^2+5x\le 6\\ & x^2+5x-6\le 0\\ & (x+6)(x-1)\le 0\end{array}\\ \mathbf{\text{Lihat Gambar 1}}& \mathbf{\text{Lihat Gambar 2}}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Nilai}x\text{yang memenuhi}\left|{\displaystyle \frac{1}{2}x+6}\right|\ge 9\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -12<x<6\\ \text{b}.& -30\le x\le 6\\ \text{c}.& x\ge 6\text{atau}x\le -30\\ \text{d}.& x<6\text{atau}x<-30\\ \text{e}.& x\le 6\text{atau}x\ge -30\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \left|{\displaystyle \frac{1}{2}x+6}\right|\ge 9\\ & {\displaystyle \frac{1}{2}x+6\le -9\text{atau}{\displaystyle \frac{1}{2}x+6\ge 9}}\\ & {\displaystyle \frac{1}{2}x\le -9-6\text{atau}{\displaystyle \frac{1}{2}x\ge 9-6}}\\ & {\displaystyle \frac{1}{2}x\le -15\text{atau}{\displaystyle \frac{1}{2}x\ge 3}}\\ & x\le -30\text{atau}x\ge 6\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & 3|x+1|\le |x-2|\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{4}\le x\le -\frac{1}{4}}\\ \text{b}.& -{\displaystyle \frac{5}{2}\le x\le \frac{5}{2}}\\ \text{c}.& x\le {\displaystyle \frac{1}{4}\text{atau}x\ge \frac{5}{2}}\\ \text{d}.& x\le -{\displaystyle \frac{5}{2}\text{atau}x\ge \frac{1}{4}}\\ \text{e}.& x\le -{\displaystyle \frac{5}{2}\text{atau}x\ge -\frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& 3|x+1|\le |x-2|\\ & {\left(3|x+1|\right)}^2\le {\left(|x-2|\right)}^2\\ & {(3x+3)}^2\le {(x-2)}^2\\ & (3x+3+(x-2))(3x+3-(x-2))\le 0\\ & (4x+1)(2x+5)\le 0\\ & \text{HP}=\{x|-{\displaystyle \frac{5}{2}\le x\le -\frac{1}{4},x\in \mathbb{R}}\}\end{array}\end{array}$.

$\begin{array}{l}\\ 31.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |x-3|<3\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<3\\ \text{b}.& -3<x<3\\ \text{c}.& x<-3\text{atau}x<3\\ \text{d}.& x>0\text{atau}x<6\\ \text{e}.& x<0\text{atau}x<6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& |x-3|<3\\ & -3<(x-3)<3\\ & -3+3<x<3+3\\ & 0<x<6\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |x+4|>8\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x>-8\\ \text{b}.& x<4\text{atau}x>12\\ \text{c}.& x>4\text{atau}x>-12\\ \text{d}.& x<-4\text{atau}x<6\\ \text{e}.& x>4\text{atau}x<-12\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& |x+4|>8\\ & (x+4)<-8\text{atau}(x+4)>8\\ & x<-12\text{atau}x>4\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Himpunan penyelesaian dari pertidaksamaan}\\ & \left|{\displaystyle \frac{x+1}{2}}\right|>\left|{\displaystyle \frac{x-2}{3}}\right|\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \text{HP}=\{x|-7<x<{\displaystyle \frac{1}{5},x\in \mathbb{R}}\}\\ \text{b}.& \text{HP}=\{x|x<-7\text{atau}x>{\displaystyle \frac{1}{5},x\in \mathbb{R}}\}\\ \text{c}.& \text{HP}=\{x|x>-7,x\in \mathbb{R}\}\\ \text{d}.& \text{HP}=\{x|-1<x<2,x\in \mathbb{R}\}\\ \text{e}.& \text{HP}=\{x|x<-1\text{atau}x>2,x\in \mathbb{R}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \left|{\displaystyle \frac{x+1}{2}}\right|>\left|{\displaystyle \frac{x-2}{3}}\right|\\ & {\left({\displaystyle \frac{x+1}{2}}\right)}^2>{\left({\displaystyle \frac{x-2}{3}}\right)}^2\\ & \left({\displaystyle \frac{x+1}{2}+\frac{x-2}{3}}\right)\left({\displaystyle \frac{x+1}{2}-{\displaystyle \frac{x-2}{3}}}\right)>0\\ & \left({\displaystyle \frac{3(x+1)+2(x-2)}{6}}\right)\left({\displaystyle \frac{3(x+1)-2(x-2)}{6}}\right)>0\\ & \left({\displaystyle \frac{5x-1}{6}}\right)\left({\displaystyle \frac{x+7}{6}}\right)>0\\ & \text{HP}=\{x|x<-7\text{atau}x>{\displaystyle \frac{1}{5},x\in \mathbb{R}}\}\end{array}\end{array}$.

$\begin{array}{l}\\ 34.& \text{Penyelesaian dari pertidaksamaan}\\ & \left|{\displaystyle \frac{3-2x}{-5}}\right|>5\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-11\text{atau}x>14\\ \text{b}.& x<-14\text{atau}x>11\\ \text{c}.& 11<x<14\\ \text{d}.& -14<x<-11\\ \text{e}.& x>14\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \left|{\displaystyle \frac{3-2x}{-5}}\right|>5\\ & {\displaystyle \frac{3-2x}{-5}<-5\text{atau}{\displaystyle \frac{3-2x}{-5}>5}}\\ & {\displaystyle \frac{2x-3}{5}>5\text{atau}{\displaystyle \frac{2x-3}{5}<-5}}\\ & 2x-3>25\text{atau}2x-3<-25\\ & 2x>25+3\text{atau}2x<-25+3\\ & x>14\text{atau}x<-11,\\ & \text{dapat juga dituliskan}\\ & x<-11\text{atau}x>14\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |2-2|x+1||>4\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-4\text{atau}x>2\\ \text{b}.& x<-3\text{atau}x>1\\ \text{c}.& x<-2\text{atau}x>0\\ \text{d}.& x<-1\text{atau}x>3\\ \text{e}.& x<0\text{atau}x>4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& |2-2|x+1||>4\\ & 2-2|x+1|<-4\text{atau}2-2|x+1|>4\\ & -2|x+1|<-6\text{atau}-2|x+1|>2\\ & |x+1|>3\text{atau}|x+1|<-1\\ & \{\begin{array}{c}(x+1)<-3\\ (x+1)>3\end{array}\text{atau}\{\begin{array}{c}|x+1|<-1\\ \mathbf{\text{tak mungkin}}\end{array}\\ & \text{Selanjutnya}\text{akan didapatkan}\\ & x<-4\text{atau}x>2\end{array}\end{array}$

Latihan Soal 3 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 16.& \text{Penyelesaian yang memenuhi untuk}\\ & |3x-(4x-7)|=6\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{-13,-1\}& & & \text{d}.& \{-13,1\}\\ \text{b}.& \{-1,13\}& \text{c}.& \{1,13\}& \text{e}.& \left\{13\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}|3x-(4x-7)|& =6\\ |3x-4x+7|& =6\\ |-x+7|& =6\\ (-x+7)& =\pm 6\\ -x+7& =\{\begin{array}{l}6\iff -x=6-7\iff x=1\\ \\ -6\iff -x=-6-7\iff x=13\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Penyelesaian yang memenuhi untuk}\\ & |x-1|=2x+1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{-2\}& & & \text{d}.& \left\{\right\}\\ \text{b}.& \{-2,0\}& \text{c}.& \{-1\}& \text{e}.& \left\{0\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& |x-1|=2x+1\\ & (x-1)=\pm (2x+1)\\ & (x-1)=\{\begin{array}{l}+(2x+1)\\ \\ -(2x+1)\end{array}\\ & \begin{array}{rl}& \\ \text{Syarat}& :\\ (x-1)& \{\begin{array}{l}x-1\ge 0\iff x\ge 1\\ \\ x-1<0\iff x<1\end{array}\end{array}\\ & \begin{array}{|cc|}\hline x\ge 1& x<1\\ \text{Proses}& \text{Proses}\\ \begin{array}{rl}(x-1)& =+(2x+1)\\ x-2x& =1+1\\ -x& =2\\ x& =-2\end{array}& \begin{array}{rl}(x-1)& =-(2x+1)\\ x+2x& =-1+1\\ 3x& =0\\ x& =0\end{array}\\ \text{tidak memenuhi}& \mathbf{\text{memenuhi}}\\ \hline\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Penyelesaian yang memenuhi untuk}\\ & |3x+1|=2x+9\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{-2\}& & \text{d}.& \left\{\right\}\\ \text{b}.& \left\{8\right\}& & \text{e}.& \text{setiap bilangan real}\\ \text{c}.& \{-2,8\}& \end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& |3x+1|=2x+9\\ & (3x+1)=\pm (2x+9)\\ & (3x+1)=\{\begin{array}{l}+(2x+9)\\ \\ -(2x+9)\end{array}\end{array}\\ & \begin{array}{rl}\text{Syarat}& :\\ (3x+1)& \{\begin{array}{l}3x+1\ge 0\iff x\ge -{\displaystyle \frac{1}{3}}\\ \\ 3x+1<0\iff x<-{\displaystyle \frac{1}{3}}\end{array}\\ & \end{array}\\ & \begin{array}{|cc|}\hline x\ge -{\displaystyle \frac{1}{3}}& x<-{\displaystyle \frac{1}{3}}\\ \text{Proses}& \text{Proses}\\ \begin{array}{rl}(3x+1)& =+(2x+9)\\ 3x-2x& =9-1\\ x& =8\\ & \end{array}& \begin{array}{rl}(3x+1)& =-(2x+9)\\ 3x+2x& =-9-1\\ 5x& =-10\\ x& =-2\end{array}\\ \mathbf{\text{memenuhi}}& \mathbf{\text{memenuhi}}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Jumlah akar-akar dari}{x}^2+\left|x\right|-6=0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& 4\\ \\ & & & (\mathbf{\text{Entrance Examination}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{x}^2+\left|x\right|-6& =0\\ (\left|x\right|+3)(\left|x\right|-2)& =0\\ \left|x\right|+3=0\text{atau}\left|x\right|-2& =0\\ \left|x\right|=-3(\mathbf{\text{tm}})\text{atau}\left|x\right|& =2(\mathbf{\text{mm}})\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}x& =\pm 2\{\begin{array}{ll}{x}_1& =2\\ x_2& =-2\end{array}\\ & \text{untuk jumlah}\text{dari akar-akarnya adalah}:\\ & x_1+x_2=2+(-2)\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Penyelesaian pertidaksamaan}{x}^2+\left|x\right|-6\le 0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2\le x<0\\ \text{b}.& 0\le x\le 2\\ \text{c}.& -2\le x\le 2\\ \text{d}.& -3\le x\le 2\\ \text{e}.& -2\le x\le 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ \\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}& \text{Proses penyelesaian dipecah jadi 2 bagian}\\ & \text{yaitu}:\{\begin{array}{ll}x& \ge 0\\ x& <0\end{array}\\ & \text{Diketahui pertidaksamaan}:x^2+\left|x\right|-6\le 0\\ & \begin{array}{|cc|}\hline (1)& (2)\\ x\ge 0& x<0\\ \text{maka}\left|x\right|=x& \text{maka}\left|x\right|=-x\\ \begin{array}{rl}& x^2+(x)-6\le 0\\ & x^2+x-6\le 0\\ & (x+3)(x-2)\le 0\\ & \text{Selesaian}:\\ & -3\le x\le 2\\ & \text{karena}x\ge 0,\\ & \text{maka penyelesaian}\\ & \text{menjadi}\\ & 0\le x\le 2\end{array}& \begin{array}{rl}& x^2+(-x)-6\le 0\\ & x^2-x-6\le 0\\ & (x+2)(x-3)\le 0\\ & \text{Selesaian}:\\ & -2\le x\le 3\\ & \text{karena}x<0,\\ & \text{maka penyelesaian}\\ & \text{menjadi}\\ & -2\le x<0\end{array}\\ \hline\end{array}\\ & \text{Gabungan dari penyelesaian (1) dan (2)}\\ & \text{adalah}:-2\le x\le 2\end{array}\\ \\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& \text{Diketahui pertidaksamaan}:x^2+\left|x\right|-6\le 0\\ & \text{dan perlu diingat pula bahwa}:\left|x\right|\ge 0\\ & \text{diubah menjadi}:{\left|x\right|}^2+\left|x\right|-6\le 0\\ & \iff (\left|x\right|+3)(\left|x\right|-2)\le 0\\ & \iff -3\le \left|x\right|\le 2\\ & \text{karena}\left|x\right|\ge 0,\text{maka}\\ & 0\le \left|x\right|\le 2\Rightarrow \left|x\right|\le 2\\ & \text{Sehingga penyelesaian menjadi}\\ & -2\le x\le 2\end{array}\end{array}$.

$\begin{array}{l}\\ 21.& \text{Seluruh bilangan bilangan real}x\\ & \text{yang jaraknya terhadap 3 kurang dari 1 }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3<x<4& \\ \text{b}.& 2<x<3\\ \text{c}.& 2<x<4\\ \text{d}.& 3<x<5\\ \text{e}.& 1<x<3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Seluruh }\text{bilangan bilangan real}x\\ & \text{yang jaraknya terhadap 3 kurang dari 1, }\\ & \text{maksudnya adalah}:\\ & |x-3|<1\\ & \iff -1<x-3<1\\ & \iff -1+\mathbf{\text{(3)}}<x-3+\mathbf{\text{(3)}}<1+\mathbf{\text{(3)}}\\ & \iff 2<x<4\end{array}\end{array}$.

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

$\begin{array}{l}\\ 23.& \text{Nilai}x\text{real yang memenuhi untuk}\\ & |2x-9|<3\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -3\le x\le 6& \\ \text{b}.& -3<x<6\\ \text{c}.& 3<x<6\\ \text{d}.& 3\le x\le 6\\ \text{e}.& -3<x<-6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& |2x-9|<3\\ & \iff -3<2x-9<3\\ & -3+(9)<2x-9+(9)<3+(9)\\ & \text{masing-masing ditambah 9}\\ & \text{dan akan menjadi bentuk}\\ & 6<2x<12\\ & 6.\left({\displaystyle \frac{1}{2}}\right)<2x.\left({\displaystyle \frac{1}{2}}\right)<12.\left({\displaystyle \frac{1}{2}}\right)\\ & \text{masing-masing dikali}{\displaystyle \frac{1}{2}}\\ & \text{dan akan berubah menjadi bentuk}\\ & 3<x<6\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Nilai}x\text{real yang memenuhi untuk}\\ & |3x+5|\ge 19\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x\le -{\displaystyle \frac{14}{3}\text{atau}x\ge 8}& \\ \\ \text{b}.& x<-8\text{atau}x>{\displaystyle \frac{14}{3}}\\ \\ \text{c}.& x\le -8\text{atau}x\ge {\displaystyle \frac{14}{3}}\\ \\ \text{d}.& x<-{\displaystyle \frac{14}{3}\text{atau}x>8}\\ \\ \text{e}.& x\le 8\text{atau}x\ge -{\displaystyle \frac{14}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& |3x+5|\ge 19\\ \\ & (\ast )-19\ge 3x+5\text{atau}& (\ast \ast )3x+5\ge 19\\ & -19-5\ge 3x\text{atau}& 3x\ge 19-5\\ & -{\displaystyle \frac{24}{3}\ge x\text{atau}}& x\ge {\displaystyle \frac{14}{3}}\\ & -8\ge x\text{atau}& x\ge {\displaystyle \frac{14}{3}}\\ & x\le -8\text{atau}& x\ge {\displaystyle \frac{14}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Nilai}x\text{real yang memenuhi}\\ & 25-|10x+5|\ge |40x-20|\text{adalah}....\\ \\ & (\mathbf{\text{NUS Entrance Examination A level}})\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}:\\ & \begin{array}{rl}& 25-|10x+5|\ge |40x-20|\\ & 25-5|2x+1|\ge 20|2x-1|\\ & 5-|2x+1|\ge 4|2x-1|\\ & \text{ilustrasinya}\begin{array}{lllllll}& & & & & & \\ & & -\frac{1}{2}& & & \frac{1}{2}& & \end{array}\\ & \text{dan berikut}\text{pembagian wilayahnya}\\ & \begin{array}{|ccc|}\hline -\mathrm{\infty }<x<-{\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\le x<\frac{1}{2}}& {\displaystyle \frac{1}{2}\le x<\mathrm{\infty }}\\ \{\begin{array}{ll}|2x+1|& =-(2x+1)\\ |2x-1|& =-(2x-1)\end{array}& \{\begin{array}{ll}|2x+1|& =+(2x+1)\\ |2x-1|& =-(2x-1)\end{array}& \{\begin{array}{ll}|2x+1|& =+(2x+1)\\ |2x-1|& =+(2x-1)\end{array}\\ \hline\end{array}\\ & \text{Selanjutnya adalah}\end{array}\end{array}$

$.\begin{array}{|ccc|}\hline \begin{array}{rl}-\mathrm{\infty }<x<& -\frac{1}{2},\\ 25-|10x+5|& \ge |40x-20|\\ 25-5|2x+1|& \ge 20|2x-1|\\ 5-(-(2x+1))& \ge 4(-(2x-1))\\ 5+2x+1& \ge -8x+4\\ 10x& \ge -2\\ x& \ge -\frac{2}{10}(\mathbf{\text{tm}})\\ & \end{array}& \begin{array}{rl}-\frac{1}{2}\le x<& \frac{1}{2},\\ 25-5|2x+1|& \ge 20|2x-1|\\ 5-(2x+1)& \ge 4(-(2x-1))\\ 5-2x-1& \ge -8x+4\\ 6x& \ge 0\\ x& \ge 0(\mathbf{\text{mm}})\\ \text{yang memenuhi}& =\{0\le x<\frac{1}{2}\}\end{array}& \begin{array}{rl}\frac{1}{2}\le x<& \mathrm{\infty },\\ 25-5|2x+1|& \ge 20|2x-1|\\ 5-(2x+1))& \ge 4(2x-1)\\ 5-2x-1& \ge 8x-4\\ -10x& \ge -8\\ x& \le \frac{8}{10}(\mathbf{\text{mm}})\\ \text{yang memenuhi}& =\{\frac{1}{2}\le x\le \frac{4}{5}\}\end{array}\\ \hline\end{array}$

$.\begin{array}{rl}& \\ & \text{Sehingga yang memenuhi}\text{adalah}:\\ & =\{0\le x\le {\displaystyle \frac{4}{5}}\}\end{array}$.

sumber soal di sini dan di sini

Latihan Soal 2 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 11.& \text{Perhatikanlah ilustrasi grafik di bawah ini}\\ & \begin{array}{}\end{array}\end{array}$.

$.\begin{array}{l}\\ & \text{Persamaan yang memenuhi rumus tersebut }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y=|-x-2|\\ \text{b}.& y=-2x-4\\ \text{c}.& y=-|2x-4|\\ \text{d}.& y=|-2x-4|\\ \text{e}.& y=|-2x+4|\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Dengan cara substitusi langsung kita}\\ & \text{akan mendapatkan}\\ & \bullet \text{untuk}x=4\text{menyebabkan nilai}y=4\\ & \text{dan sampai langkah di sini hanya ada }\\ & \text{1 persamaan yang memenuhi yaitu}:\\ & y=|-2x+4|\end{array}\end{array}$.

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

$.\begin{array}{l}\\ & \text{Perhatikanlah ilustrasinya grafik di bawah ini}\\ & \begin{array}{}\end{array}\end{array}$.

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

$.\begin{array}{r}\begin{array}{rl}& \text{untuk}:x>0,y>0\mathbf{\text{(kuadran I)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (x)+x+y=5\\ & \iff 2x+y=5(\text{ada})\\ x+\left|y\right|-y=10& \iff x+(y)-y=10\\ & \iff x=10(\text{ada})\end{array}\\ & \text{untuk}:x<0,y>0\mathbf{\text{(kuadran II)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (-x)+x+y=5\\ & \iff y=5(\text{ada})\\ x+\left|y\right|-y=10& \iff x+(y)-y=10\\ & \iff x=10\\ & \mathbf{\text{(tidak memenuhi)}}\end{array}\end{array}\end{array}$.

$.\begin{array}{r}\begin{array}{rl}& \text{untuk}:x<0,y<0\mathbf{\text{(kuadran III)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (-x)+x+y=5\\ & \iff y=5\\ & \mathbf{\text{(tidak memenuhi)}}\\ x+\left|y\right|-y=10& \iff x+(-y)-y=10\\ & \iff x-2y=10(\text{ada})\end{array}\\ & \text{untuk}:x>0,y<0\mathbf{\text{(kuadran IV)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (x)+x+y=5\\ & \iff 2x+y=5(\text{ada})\\ x+\left|y\right|-y=10& \iff x+(-y)-y=10\\ & \iff x-2y=10(\text{ada})\end{array}\end{array}\end{array}$.

$.\begin{array}{l}\\ & \text{Sebagai ilustrasinya, berikut grafiknya}\\ & \begin{array}{}\end{array}\end{array}$.

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

$.\begin{array}{l}\\ & \mathbf{\text{Jawab}}:\\ & \text{Berikut ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{a}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan berikut ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{b}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan berikut pula ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{c}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan berikut pula ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{d}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan sebagai ilustrasinya}\\ & \text{lihat soal no}\mathbf{\text{c}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan terakhir berikut ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{f}}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Penyelesaian yang memenuhi untuk}\\ & |3x-(4x-7)|=6\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{-13,-1\}& & & \text{d}.& \{-13,1\}\\ \text{b}.& \{-1,13\}& \text{c}.& \{1,13\}& \text{e}.& \left\{13\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}|3x-(4x-7)|& =6\\ |3x-4x+7|& =6\\ |-x+7|& =6\\ (-x+7)& =\pm 6\\ -x+7& =\{\begin{array}{l}6\iff -x=6-7\iff x=1\\ \\ -6\iff -x=-6-7\iff x=13\end{array}\end{array}\end{array}$.

Latihan Soal 1 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 1.& \text{Nilai untuk}-|-2021|=....\\ & \begin{array}{l}\\ \text{a}.& -2021& & & \text{d}.& -{2021}^{-1}\\ \text{b}.& 2021& \text{c}.& {2021}^{-1}& \text{e}.& {2021}^2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}-|-2021|& =-\left(2021\right)=-2021\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Nilai untuk}|-4|-|-6^2\times 2|=....\\ & \begin{array}{l}\\ \text{a}.& -68& & & \text{d}.& 68\\ \text{b}.& -40& \text{c}.& 40& \text{e}.& 76\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}|-4|-|-6^2\times 2|& =\left(4\right)-|-36\times 2|\\ & =4-|-72|\\ & =4-\left(72\right)\\ & =-68\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Nilai untuk}(-2022)-|-(-2021)|-{|-3^2|}^2=....\\ & \begin{array}{l}\\ \text{a}.& -3960& & & \text{d}.& -4068\\ \text{b}.& -4038& \text{c}.& -4050& \text{e}.& -4124\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& (-2022)-|-(-2021)|-{|-3^2|}^2\\ & =-2022-2021-{|-9|}^2\\ & =-4043-9^2\\ & =-4043-81\\ & =-4124\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai untuk}{|-4|}^{|-2|}-{|-2|}^{|-4|}=....\\ & \begin{array}{l}\\ \text{a}.& -32& & & \text{d}.& 16\\ \text{b}.& -16& \text{c}.& 0& \text{e}.& 32\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{|-4|}^{|-2|}-{|-2|}^{|-4|}& =4^2-2^4\\ & =16-16\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai untuk}\\ & |1-2^2|+|2^2-3^2|+|3^2-4^2|+\cdots +|{2020}^2-{2021}^2|....\\ & \begin{array}{l}\\ \text{a}.& -{2021}^2& & & \text{d}.& {2021}^2-1\\ \text{b}.& 1-{2021}^2& \text{c}.& -2021& \text{e}.& {2021}^2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& |1-2^2|+|2^2-3^2|+|3^2-4^2|+\cdots +|{2020}^2-{2021}^2|\\ & =(2^2-1)+(3^2-2^2)+(4^2-3^2)+\cdots +({2021}^2-{2020}^2)\\ & =-1+{2021}^2\\ & ={2021}^2-1\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Nilai untuk}\\ & \left|{\displaystyle \frac{1}{2}-1}\right|\times \left|{\displaystyle \frac{1}{3}-1}\right|\times \left|{\displaystyle \frac{1}{4}-1}\right|\times \left|{\displaystyle \frac{1}{5}-1}\right|\times \cdots \times \left|{\displaystyle \frac{1}{2021}-1}\right|=....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2021}}& & & \text{d}.& {\displaystyle \frac{2020}{2021}}\\ \\ \text{b}.& -{\displaystyle \frac{2020}{2021}}& \text{c}.& -2021{\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle \frac{1}{2021}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \left|{\displaystyle \frac{1}{2}-1}\right|\times |\frac{1}{3}-1|\times |\frac{1}{4}-1|\times |\frac{1}{5}-1|\times \cdots \times |\frac{1}{2021}-1|\\ & =|-{\displaystyle \frac{1}{2}}|\times |-\frac{2}{3}|\times |-\frac{3}{4}|\times |-\frac{4}{5}|\times \cdots \times |-\frac{2020}{2021}|\\ & ={\displaystyle \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \cdots \times \frac{2019}{2020}\times \frac{2020}{2021}}\\ & ={\displaystyle \frac{1}{2021}}\end{array}\end{array}$.

$\begin{array}{ll}7.& \text{Bentuk sederhana dari}x-5y\text{dan}\\ & 5y-x\text{adalah}....\\ & \begin{array}{llll}\text{a}.& |5y-x|& \text{d}.& |y-x|\\ \text{b}.& |y-5x|& \text{e}.& |-5y-x|\\ \text{c}.& |x-5y|\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & |x-5y|=m=\{\begin{array}{ll}\bullet & =x-5y\\ \bullet & =-(x-5y)\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Nilai}m\text{yang memenuhi}\left|4m\right|=16\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 4\\ \text{c}.& \pm 2\\ \text{d}.& \pm 4\\ \text{e}.& \pm 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left|4m\right|& =16\\ (4m)& =\pm 16\\ m& =\pm {\displaystyle \frac{16}{4}}\\ & =\pm 4\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Nilai}x\text{yang memenuhi untuk}|2x+5|=9\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2& & \text{d}.& -7\text{dan}2\\ \text{b}.& 2\text{dan}7& & \text{e}.& -2\text{dan}7\\ \text{c}.& -7\text{dan}-2& \end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}|2x+5|& =9\\ (2x+5)& =\pm 9\\ 2x& =\pm 9-5\\ x& ={\displaystyle \frac{\pm 9-5}{2}}\\ x& =\{\begin{array}{l}={\displaystyle \frac{+9-5}{2}=\frac{4}{2}=2}\\ \\ ={\displaystyle \frac{-9-5}{2}=\frac{-14}{2}=-7}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Nilai}x\text{yang memenuhi}10-4|4-5x|=26\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\text{atau}1{\displaystyle \frac{2}{3}}& & \text{d}.& 1\text{atau}-{\displaystyle \frac{3}{5}}\\ \\ \text{b}.& 2\text{atau}2{\displaystyle \frac{3}{5}}& & \text{e}.& -1\text{atau}2{\displaystyle \frac{3}{5}}\\ \text{c}.& -2{\displaystyle \frac{3}{4}\text{atau}1}& \end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}10-4|4-5x|& =-26\\ -4|4-5x|& =-36\\ |4-5x|& =9\\ (4-5x)& =\pm 9\\ -5x& =-4\pm 9\\ x& ={\displaystyle \frac{-4\pm 9}{-5}}\\ x& =\{\begin{array}{ll}{\displaystyle \frac{-4+9}{-5}}& =-1\\ \text{atau}& \\ {\displaystyle \frac{-4-9}{-5}}& ={\displaystyle \frac{13}{5}=2\frac{3}{5}}\end{array}\end{array}\end{array}$