Contoh Soal 1 Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

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

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

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

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

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