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

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

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

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

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

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