$\begin{array}{ll}\\ 11.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{5}{4x-3} \right |\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\leq x\leq \frac{3}{4}\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{b}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \displaystyle \frac{3}{4}< x\leq 2\\ \textrm{c}.&-\displaystyle \frac{1}{2}\leq x\leq 2,\: \: x\neq \frac{3}{4}\\ \textrm{d}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x>\frac{3}{4}\\ \color{red}\textrm{e}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x\geq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{5}{4x-3} \right |&\leq 1\\ -1\leq \displaystyle \frac{5}{4x-3}&\leq 1,\: \: \color{magenta}\textbf{jika dibalik}\\ -1\geq \displaystyle \frac{4x-3}{5}&\geq 1,\: \: \color{magenta}\textbf{bentuk ini tidak}\\ \color{magenta}\textbf{dibolehkan}&\: \color{magenta}\textbf{maka perlu diubah menjadi}\\ -1\geq \displaystyle \frac{4x-3}{5}\: \: \textrm{atau}&\: \: \displaystyle \frac{4x-3}{5}\geq 1,\: \: \color{black}\textrm{selanjutnya}\\ \bullet \quad \textrm{bagian}&\: 1\\ -1&\geq \displaystyle \frac{4x-3}{5}\Leftrightarrow \frac{4x-3}{5}\leq -1\\ 4x-3&\leq -5\\ 4x&\leq -2\\ x&\leq -\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{4x-3}{5}&\geq 1\\ 4x-3&\geq 5\\ 4x&\geq 8\\ x&\geq 2 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 12.&(\textrm{UMPTN 95})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2}{2x-1} \right |> 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x> 2\\ \textrm{b}.&x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{c}.&x<-1\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{d}.&-1<x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{e}.&x<-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{semua opsi bukan jawaban}\\ &\textbf{Berikut pembahasannya}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{2}{2x-1} \right |&> 1\\ -1>\displaystyle \frac{2}{2x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2}{2x-1}>1,\: \color{magenta}\textbf{dibalik}\\ -1<\displaystyle \frac{2x-1}{2}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x-1}{2}<1\\ \bullet \quad \textrm{bagian}&\: 1\\ \displaystyle \frac{2x-1}{2}&>-1\\ 2x-1&>-2\\ 2x&>-1\\ x&>-\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{2x-1}{2}&<1\\ 2x-1&<2\\ 2x&<3\\ x&<\displaystyle \frac{3}{2} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 13.&(\textrm{UMPTN 00})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2x+7}{x-1} \right |\geq 1\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-2\leq x\leq 8\\ \textrm{b}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -2\\ \textrm{c}.&-8\leq x< 1\: \: \textrm{atau}\: \: x>1\\ \textrm{d}.&-2\leq x< 1\: \: \textrm{atau}\: \: 1< x\leq 8\\ \color{red}\textrm{e}.&x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{2x+7}{x-1} \right |&\geq 1\\ -1\geq \displaystyle \frac{2x+7}{x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x+7}{x-1}\geq 1\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{2x+7}{x-1}&\leq -1\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{2x+7}{x-1}&+1\leq 0\\ &\displaystyle \frac{2x+7+(x-1)}{x-1}\leq 0\\ \displaystyle \frac{3x+6}{x-1}&\leq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| -2\leq x< 1,\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{2x+7}{x-1}&\geq 1\\ \displaystyle \frac{2x+7}{x-1}&-1\geq 0\\ &\displaystyle \frac{2x+7-(x-1)}{x-1}\geq 0\\ \displaystyle \frac{x+8}{x-1}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>1,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x> 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 14.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{x-2}{x+3} \right |\leq 2\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-8\leq x< -3\\ \textrm{b}.&-8\leq x< -1\\ \textrm{c}.&-4\leq x< -3\\ \color{red}\textrm{d}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3}\\ \textrm{e}.&x\leq -4\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{x-2}{x+3} \right |&\leq 2\\ -2\leq \displaystyle \frac{x-2}{x+3}&\: \: \textrm{atau}\: \: \displaystyle \frac{x+2}{x+3}\leq 2\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{x-2}{x+3}&\geq -2\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{x-2}{x+3}&+2\geq 0\\ &\displaystyle \frac{x-2+2(x+3)}{x+3}\geq 0\\ \displaystyle \frac{3x+4}{x+3}&\geq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| x< -3\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{x-2}{x+3}&\leq 2\\ \displaystyle \frac{x-2}{x+3}&-2\leq 0\\ &\displaystyle \frac{x-2-2(x+3)}{x+3}\leq 0\\ \displaystyle \frac{-x-8}{x+3}&\leq 0,\: \: \color{magenta}\textbf{koefisien \textit{x} negatif}\\ \displaystyle \frac{x+8}{x+3}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>-3,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 15.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{2}{x+1}\leq \left | x \right |\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{b}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: 0\leq x\leq 1 \right \}\\ \textrm{c}.&\left \{ x|x\geq 1 \right \}\\ \color{red}\textrm{d}.&\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{e}.&\left \{ x|-1< x\leq 1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left | x \right |&\geq \displaystyle \frac{2}{x+1}\quad\quad\quad \color{black}\textrm{berakibat}\\ \displaystyle \frac{-2}{x+1}\geq x&\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{2}{x+1}\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ x\leq \displaystyle \frac{-2}{x+1}&\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ x+\displaystyle \frac{2}{x+1}&\leq 0\\ &\displaystyle \frac{x(x+1)+2}{x+1}\leq 0\\ \displaystyle \frac{x^{2}+x+2}{x+1}&\leq 0\Leftrightarrow \displaystyle \frac{\textrm{Definit positif}}{x+1}\leq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| x< -1,\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ x&\geq \displaystyle \frac{2}{x+3}\\ x-&\displaystyle \frac{2}{x+1}\geq 0\\ &\displaystyle \frac{x(x+1)-2}{x+1}\geq 0\\ &\displaystyle \frac{x^{2}+x-2}{x+1}\geq 0\\ &\displaystyle \frac{(x+2)(x-1)}{x+1}\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|-2\leq x< -1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$
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