Latihan Soal 4 Persiapan PAS Gasal Matematika Wajib Kelas X

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

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

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

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

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

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

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

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

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

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