$\begin{array}{ll}\\ 26.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & (x+3)(x-1)\geq (x-1)\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&1\leq x\leq 3&\\ \textrm{b}.&x\leq -2\: \: \textrm{atau}\: \: x\geq 1\\ \textrm{c}.&-3\leq x\leq -1\\ \textrm{d}.&-2\geq x\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{e}.&-1\geq x\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}(x+3)(x-1)&\geq (x-1)\\ (x+3)(x-1)-(x-1)&\geq 0\\ (x-1)\left ( (x+3)-1 \right )&\geq 0\\ (x-1)(x+2)&\geq 0\\ \end{aligned}\\ &\textrm{Sehingga solusinya adalah:}\\ &\color{red}x\leq -2\: \: \color{black}\textrm{atau}\: \: \color{red}x\geq 1 \end{array}$.
$\begin{array}{ll}\\ 27.&\textrm{Himpunan penyelesaian pertidaksamaan}\\ & \left | x+3 \right |<2\left | x-4 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&\left \{ x|x<-\displaystyle \frac{5}{3} \right \}\\ \textrm{b}.&\left \{ x|\: \displaystyle \frac{5}{3}<x<-11 \right \}\\ \textrm{c}.&\left \{ x|x\geq -11 \right \} \\ \textrm{d}.&\left \{ x|x<\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x>11 \right \}\\ \textrm{e}.&\left \{ x|x>-\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x<-11 \right \} \\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | x+3 \right |<2\left | x-4 \right |\\ &\left ( x+3 \right )^{2}<2^{2}\left ( x-4 \right )^{2}\\ & \textrm{dikuadratkan masing-masing ruas}\\ &x^{2}+6x+9<4\left ( x^{2}-8x+16 \right )\\ &x^{2}-4x^{2}+6x+32x+9-64<0\\ &-3x^{2}+38x-55<0\\ &3x^{2}-38x+55>0\\ &\left ( 3x-5 \right )\left (x -11 \right )>0\\\\ &\textrm{Berikut untuk}\: \textrm{garis bilangannya} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 28.&\textrm{Himpunan penyelesaian pertidaksamaan}\\ & \left | x^{2}+5x \right |\leq 6\: \: \textrm{adalah... .}\\ &\textrm{a}.\quad \left \{ x|-6\leq x\leq 1 \right \}\\ &\textrm{b}.\quad \left \{ x|-3\leq x\leq -2 \right \}\\ &\textrm{c}.\quad \left \{ x|-6\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 1\right \}\\ &\textrm{d}.\quad \left \{ x|-6\leq x\leq -5 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\ &\textrm{e}.\quad \left \{ x|-5\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diketahui bahwa}\\ &\\ \left | x^{2}+5x \right |&\leq 6\\ -6\leq x^{2}+5x&\leq 6\\ & \end{aligned}\\ &\begin{array}{|c|c|}\hline \begin{aligned}&-6\leq x^{2}+5x\\ &x^{2}+5x+6\geq 0\\ &(x+3)(x+2)\geq 0 \end{aligned}&\begin{aligned}&x^{2}+5x\leq 6\\ &x^{2}+5x-6\leq 0\\ &(x+6)(x-1)\leq 0 \end{aligned}\\\hline \textbf{Lihat Gambar 1}&\textbf{Lihat Gambar 2}\\\hline \end{array} \end{array}$.
$\begin{array}{ll}\\ 29.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \left | \displaystyle \frac{1}{2}x+6 \right |\geq 9\\ & \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-12< x< 6\\ \textrm{b}.&-30\leq x\leq 6\\ \color{red}\textrm{c}.&x\geq 6\: \: \textrm{atau}\: x\leq -30\\ \textrm{d}.&x<6\: \: \textrm{atau}\: \: x<-30\\ \textrm{e}.&x\leq 6\: \: \textrm{atau}\: \: x\geq -30\end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | \displaystyle \frac{1}{2}x+6 \right |\geq 9\\ &\displaystyle \frac{1}{2}x+6\leq -9\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}x+6\geq 9\\ &\displaystyle \frac{1}{2}x\leq -9-6\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}x\geq 9-6\\ &\displaystyle \frac{1}{2}x\leq -15\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}x\geq 3\\ &\color{red}x\leq -30\: \: \color{black}\textrm{atau}\: \: \color{red}x\geq 6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 30.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &3\left | x+1 \right |\leq \left | x-2 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\leq x\leq -\frac{1}{4}\\ \color{red}\textrm{b}.&-\displaystyle \frac{5}{2}\leq x\leq \frac{5}{2}\\ \textrm{c}.&x\leq \displaystyle \frac{1}{4}\: \: \textrm{atau}\: x\geq \frac{5}{2}\\ \textrm{d}.&x\leq -\displaystyle \frac{5}{2}\: \: \textrm{atau}\: x\geq \frac{1}{4}\\ \textrm{e}.&x\leq -\displaystyle \frac{5}{2}\: \: \textrm{atau}\: x\geq -\frac{1}{4}\end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&3\left | x+1 \right |\leq \left | x-2 \right |\\ &\left (3\left | x+1 \right | \right )^{2}\leq \left (\left | x-2 \right | \right )^{2}\\ &\left ( 3x+3 \right )^{2}\leq \left (x-2 \right )^{2}\\ &(3x+3+(x-2))(3x+3-(x-2))\leq 0\\ &(4x+1)(2x+5)\leq 0\\ &\textrm{HP}=\color{red}\left \{ x|-\displaystyle \frac{5}{2}\leq x\leq -\frac{1}{4},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.
$\begin{array}{l}\\ 31.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x-3 \right |<3\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<3\\ \textrm{b}.&-3<x<3\\ \textrm{c}.&x<-3\: \: \textrm{atau}\: x<3\\ \color{red}\textrm{d}.&x>0\: \: \textrm{atau}\: x<6\\ \textrm{e}.&x<0\: \: \textrm{atau}\: x<6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | x-3 \right |<3\\ &-3<(x-3)<3\\ &-3+3<x<3+3\\ &\color{red}0<x<6 \end{aligned} \end{array}$
$\begin{array}{l}\\ 32.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x+4 \right |>8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x>-8\\ \textrm{b}.&x<4\: \: \textrm{atau}\: x>12\\ \textrm{c}.&x>4\: \: \textrm{atau}\: x>-12\\ \textrm{d}.&x<-4\: \: \textrm{atau}\: x<6\\ \color{red}\textrm{e}.&x>4\: \: \textrm{atau}\: x<-12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | x+4 \right |>8\\ &(x+4)<-8\: \: \textrm{atau}\: \: (x+4)>8\\ &\color{red}x<-12\: \: \color{black}\textrm{atau}\: \: \color{red}x>4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 33.&\textrm{Himpunan penyelesaian dari pertidaksamaan}\\ &\left | \displaystyle \frac{x+1}{2} \right |>\left | \displaystyle \frac{x-2}{3} \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{HP}=\left \{ x|-7<x<\displaystyle \frac{1}{5},\: x\in \mathbb{R} \right \}\\ \color{red}\textrm{b}.&\textrm{HP}=\left \{ x|x<-7\: \: \textrm{atau}\: \: x>\displaystyle \frac{1}{5},\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\textrm{HP}=\left \{ x|x>-7,\: x\in \mathbb{R} \right \}\\ \textrm{d}.&\textrm{HP}=\left \{ x|-1<x<2,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\textrm{HP}=\left \{ x|x<-1\: \: \textrm{atau}\: \: x>2,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\left | \displaystyle \frac{x+1}{2} \right |>\left | \displaystyle \frac{x-2}{3} \right |\\ &\left ( \displaystyle \frac{x+1}{2} \right )^{2}>\left ( \displaystyle \frac{x-2}{3} \right )^{2}\\ &\left ( \displaystyle \frac{x+1}{2}+\frac{x-2}{3} \right )\left ( \displaystyle \frac{x+1}{2}-\displaystyle \frac{x-2}{3} \right )>0\\ &\left ( \displaystyle \frac{3(x+1)+2(x-2)}{6} \right )\left ( \displaystyle \frac{3(x+1)-2(x-2)}{6} \right )>0\\ &\left ( \displaystyle \frac{5x-1}{6} \right )\left ( \displaystyle \frac{x+7}{6} \right )>0\\ &\textrm{HP}=\color{red}\left \{ x|x<-7\: \: \color{black}\textrm{atau}\: \: \color{red}x>\displaystyle \frac{1}{5},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 34.&\textrm{Penyelesaian dari pertidaksamaan}\\ &\left | \displaystyle \frac{3-2x}{-5} \right |>5\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-11\: \: \textrm{atau}\: x>14\\ \textrm{b}.&x<-14\: \: \textrm{atau}\: x>11\\ \textrm{c}.&11<x<14\\ \textrm{d}.&-14<x<-11\\ \textrm{e}.&x>14 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\left | \displaystyle \frac{3-2x}{-5} \right |>5\\ &\displaystyle \frac{3-2x}{-5}<-5\: \: \textrm{atau}\: \: \displaystyle \frac{3-2x}{-5}>5\\ &\displaystyle \frac{2x-3}{5}>5\: \: \textrm{atau}\: \: \displaystyle \frac{2x-3}{5}<-5\\ &2x-3>25\: \: \textrm{atau}\: \: 2x-3<-25\\ &2x>25+3\: \: \textrm{atau}\: \: 2x<-25+3\\ &x>14\: \: \textrm{atau}\: \: x<-11,\\ &\textrm{dapat juga dituliskan}\\ &\color{red}x<-11\: \: \color{black}\textrm{atau}\: \: \color{red}x>14 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 35.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 2-2\left | x+1 \right | \right |>4\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-4\: \: \textrm{atau}\: x>2\\ \textrm{b}.&x<-3\: \: \textrm{atau}\: x>1\\ \textrm{c}.&x<-2\: \: \textrm{atau}\: x>0\\ \textrm{d}.&x<-1\: \: \textrm{atau}\: x>3\\ \textrm{e}.&x<0\: \: \textrm{atau}\: x>4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\left | 2-2\left | x+1 \right | \right |>4\\ &2-2\left | x+1 \right |<-4\: \: \textrm{atau}\: \: 2-2\left | x+1 \right |>4\\ &-2\left | x+1 \right |<-6\: \: \textrm{atau}\: \: -2\left | x+1 \right |>2\\ &\left | x+1 \right |>3\: \: \textrm{atau}\: \: \left | x+1 \right |<-1\\ &\left\{\begin{matrix} (x+1)<-3\\ (x+1)>3 \end{matrix}\right.\: \: \textrm{atau}\: \: \left\{\begin{matrix} \left | x+1 \right |<-1\\ \color{red}\textbf{tak mungkin} \end{matrix}\right.\\ &\textrm{Selanjutnya}\: \textrm{akan didapatkan}\\ &\color{red}x<-4\: \: \color{black}\textrm{atau}\: \: \color{red}x>2 \end{aligned} \end{array}$
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