$\begin{array}{ll}\\ 16.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | 3x-(4x-7) \right |=6\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -13,-1 \right \}&&&\textrm{d}.&\left \{ -13,1 \right \}\\ \textrm{b}.&\left \{ -1,13 \right \}&\textrm{c}.&\left \{ 1,13 \right \}&\textrm{e}.&\left \{ 13 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\left | 3x-(4x-7) \right |&=6\\ \left | 3x-4x+7 \right |&=6\\ \left | -x+7 \right |&=6\\ \left ( -x+7 \right )&=\pm 6\\ -x+7&=\begin{cases} 6 \Leftrightarrow -x=6-7\Leftrightarrow x=\color{red}1\\\\ -6\Leftrightarrow -x=-6-7\Leftrightarrow x=\color{red}13 \end{cases} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 17.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | x-1 \right |=2x+1\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ -2,0 \right \}&\textrm{c}.&\left \{ -1 \right \}&\textrm{e}.&\left \{ 0 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | x-1 \right |=2x+1\\ &(x-1)=\pm (2x+1)\\ &(x-1)= \begin{cases} +(2x+1) \\\\ -(2x+1) \end{cases}\\ &\begin{aligned}&\\ \color{blue}\textrm{Syarat}&:\\ (x-1)&\begin{cases} x-1\geq 0 \Leftrightarrow x\geq 1\\\\ x-1<0 \Leftrightarrow x<1 \end{cases}\\ \end{aligned}\\ &\begin{array}{|c|c|}\hline x\geq 1&x< 1\\\hline \textrm{Proses}&\textrm{Proses}\\\hline \begin{aligned}(x-1)&=+(2x+1)\\ x-2x&=1+1\\ -x&=2\\ x&=-2 \end{aligned}&\begin{aligned}(x-1)&=-(2x+1)\\ x+2x&=-1+1\\ 3x&=0\\ x&=0 \end{aligned}\\\hline \textrm{tidak memenuhi}&\color{red}\textbf{memenuhi}\\\hline \end{array} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 18.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | 3x+1 \right |=2x+9\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ 8 \right \}&&\textrm{e}.&\textrm{setiap bilangan real}\\ \textrm{c}.&\left \{ -2,8 \right \}& \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | 3x+1 \right |=2x+9\\ &(3x+1)=\pm (2x+9)\\ &(3x+1)= \begin{cases} +(2x+9) \\\\ -(2x+9) \end{cases}\\ \end{aligned}\\ &\begin{aligned} \color{blue}\textrm{Syarat}\: \: &:\\ (3x+1)&\begin{cases} 3x+1\geq 0 \Leftrightarrow x\geq -\displaystyle \frac{1}{3}\\\\ 3x+1<0 \Leftrightarrow x<-\displaystyle \frac{1}{3} \end{cases}\\ & \end{aligned} \\ &\begin{array}{|c|c|}\hline x\geq -\displaystyle \frac{1}{3}&x< -\displaystyle \frac{1}{3}\\\hline \textrm{Proses}&\textrm{Proses}\\\hline \begin{aligned}(3x+1)&=+(2x+9)\\ 3x-2x&=9-1\\ x&=8\\ & \end{aligned}&\begin{aligned}(3x+1)&=-(2x+9)\\ 3x+2x&=-9-1\\ 5x&=-10\\ x&=-2 \end{aligned}\\\hline \color{red}\textbf{memenuhi}&\color{red}\textbf{memenuhi}\\\hline \end{array} \end{array}$.
$\begin{array}{ll}\\ 19.&\textrm{Jumlah akar-akar dari}\: \: x^{2}+\left | x \right |-6=0\\ & \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-1\\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&4\\\\ &&&(\textbf{Entrance Examination}) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}x^{2}+\left | x \right |-6&=0\\ (\left | x \right |+3)(\left | x \right |-2)&=0\\ \left | x \right |+3=0\quad \textrm{atau}\quad \left | x \right |-2&=0\\ \left | x \right |=-3\: (\textbf{tm})\quad \textrm{atau}\quad \left | x \right |&=2\: (\textbf{mm})\\ \end{aligned} \\ &\textrm{Selanjutnya}\\ &\begin{aligned} x&=\pm 2\begin{cases} x_{1}&=2 \\ x_{2} &=-2 \end{cases}\\ &\textrm{untuk jumlah}\: \textrm{dari akar-akarnya adalah}:\\ &x_{1}+x_{2}=2+(-2)\\ &=\color{red}0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 20.&\textrm{Penyelesaian pertidaksamaan}\: \: x^{2}+\left | x \right |-6\leq 0\\ & \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-2\leq x< 0\\ \textrm{b}.&0\leq x\leq 2\\ \textrm{c}.&-2\leq x\leq 2\\ \textrm{d}.&-3\leq x\leq 2\\ \textrm{e}.&-2\leq x\leq 3\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\\\ &\color{blue}\textbf{Alternatif 1}\\ &\begin{aligned}&\textrm{Proses penyelesaian dipecah jadi 2 bagian}\\ &\textrm{yaitu}:\begin{cases} x & \geq 0 \\ x & <0 \end{cases}\\ &\textrm{Diketahui pertidaksamaan}:\: x^{2}+\left | x \right |-6\leq 0\\ &\begin{array}{|c|c|}\hline (1)&(2)\\\hline x\geq 0&x<0\\\hline \textrm{maka}\: \: \left | x \right |=x&\textrm{maka}\: \: \left | x \right |=-x\\\hline \begin{aligned}&x^{2}+(x)-6\leq 0\\ &x^{2}+x-6\leq 0\\ &(x+3)(x-2)\leq 0\\ &\color{red}\textrm{Selesaian}:\\ &-3\leq x\leq 2\\ &\textrm{karena}\quad x\geq 0,\\ &\textrm{maka penyelesaian}\\ &\textrm{menjadi}\\ &\color{blue}0\leq x\leq 2 \end{aligned}&\begin{aligned}&x^{2}+(-x)-6\leq 0\\ &x^{2}-x-6\leq 0\\ &(x+2)(x-3)\leq 0\\ &\color{red}\textrm{Selesaian}:\\ &-2\leq x\leq 3\\ &\textrm{karena}\quad x<0,\\ &\textrm{maka penyelesaian}\\ &\textrm{menjadi}\\ &\color{blue}-2\leq x<0 \end{aligned}\\\hline \end{array}\\ &\textrm{Gabungan dari penyelesaian (1) dan (2)}\\ &\textrm{adalah}:\quad \color{red}-2\leq x\leq 2 \end{aligned}\\\\ &\color{blue}\textbf{Alternatif 2}\\ &\begin{aligned}&\textrm{Diketahui pertidaksamaan}:\: x^{2}+\left | x \right |-6\leq 0\\ &\textrm{dan perlu diingat pula bahwa}:\quad \color{red}\left | x \right |\geq 0\\ &\textrm{diubah menjadi}:\quad \left | x \right |^{2}+\left | x \right |-6\leq 0\\ &\Leftrightarrow (\left | x \right |+3)(\left | x \right |-2)\leq 0\\ &\Leftrightarrow -3\leq \left | x \right |\leq 2\\ &\textrm{karena}\quad \left | x \right |\geq 0,\: \textrm{maka}\\ &0\leq \left | x \right |\leq 2\Rightarrow \left | x \right |\leq 2\\ &\textrm{Sehingga penyelesaian menjadi}\\ &\color{red}-2\leq x\leq 2 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 21.&\textrm{Seluruh bilangan bilangan real}\: \: x\\ &\textrm{yang jaraknya terhadap 3 kurang dari 1 }\\ &\textrm{adalah}\: ... .\\ &\begin{array}{lllllll}\\ \textrm{a}.&3<x<4&\\ \textrm{b}.&2<x<3 \\ \textrm{c}.&2<x<4 \\ \textrm{d}.&3<x<5\\ \textrm{e}.&1<x<3\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Seluruh }\textrm{bilangan bilangan real}\: \: x\\ &\textrm{yang jaraknya terhadap 3 kurang dari 1, }\\ &\textrm{maksudnya adalah}:\\ &\left | x-3 \right |<1\\ &\Leftrightarrow -1<x-3<1\\ &\Leftrightarrow -1+\textbf{(3)}<x-3+\textbf{(3)}<1+\textbf{(3)}\\ &\Leftrightarrow \color{red}2<x<4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 22.&\textrm{Pernyataan berikut yang tepat adalah}\: ... .\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{b}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<1\\ \textrm{c}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & -3<m<-2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{d}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & 2<m<3\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{e}.&\textrm{pilihan jawaban baik a, b, c,}\\ &\textrm{maupun d tidak ada yang benar} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikanlah opsi}\: \: \textbf{a}\: ,\\ \left | m \right |&<2\\ -2<m&<2\\ \textrm{sehing}&\textrm{ga untuk nilai}\: \: m\in \mathbb{R}\: \: \textrm{pada rentang}\\ -1<m&<2\: \: \: \: \textrm{akan memenuhi semua} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 23.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\\ & \left | 2x-9 \right |< 3\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-3\leq x\leq 6&\\ \textrm{b}.&-3<x<6\\ \textrm{c}.&3<x<6\\ \textrm{d}.&3\leq x\leq 6\\ \textrm{e}.&-3<x<-6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | 2x-9 \right |< 3\\ &\Leftrightarrow -3<2x-9<3\\ &-3+(9)<2x-9+(9)<3+(9)\\ &\textnormal{masing-masing ditambah 9}\\ &\textnormal{dan akan menjadi bentuk}\\ &6<2x<12\\ &\color{blue}6.\left ( \displaystyle \frac{1}{2} \right )<2x.\left ( \displaystyle \frac{1}{2} \right )<12.\left ( \displaystyle \frac{1}{2} \right )\\ &\textnormal{masing-masing dikali}\: \: \displaystyle \frac{1}{2}\\ &\textnormal{dan akan berubah menjadi bentuk}\\ &\color{red}3<x<6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 24.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\\ & \left | 3x+5 \right |\geq 19\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&x\leq -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x\geq 8&\\\\ \textrm{b}.&x<-8\: \: \textrm{atau}\: \: x>\displaystyle \frac{14}{3}\\\\ \textrm{c}.&x\leq -8\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{14}{3}\\\\ \textrm{d}.&x< -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x> 8\\\\ \textrm{e}.&x\leq 8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{14}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | 3x+5 \right |\geq 19\\\\ &(\ast )-19\geq 3x+5\quad \textrm{atau}&(\ast \ast )\: \: 3x+5\geq 19\\ &-19-5\geq 3x\quad \textrm{atau}&3x\geq 19-5\\ &-\displaystyle \frac{24}{3}\geq x\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ &-8\geq x\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ &x\leq \color{red}-8\quad \color{black}\textrm{atau}&x\color{red}\geq \displaystyle \frac{14}{3} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 25.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi}\\ & 25-\left | 10x+5 \right |\geq \left | 40x-20 \right |\: \: \textrm{adalah}... .\\\\ &\qquad (\textbf{NUS Entrance Examination A level})\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}:\\ &\begin{aligned}&25-\left | 10x+5 \right |\geq \left | 40x-20 \right |\\ &25-5\left | 2x+1 \right |\geq 20\left | 2x-1 \right |\\ &5-\left | 2x+1 \right |\geq 4\left | 2x-1 \right |\\ &\color{red}\textrm{ilustrasinya}\quad \color{black}\begin{array}{llllllllll} &&&&&&\\\hline &&-\frac{1}{2}&&&\frac{1}{2}&& \end{array}\\ &\textrm{dan berikut}\: \textrm{pembagian wilayahnya}\\ &\begin{array}{|c|c|c|}\hline -\infty < x< -\displaystyle \frac{1}{2}&-\displaystyle \frac{1}{2}\leq x< \frac{1}{2}&\displaystyle \frac{1}{2}\leq x< \infty \\\hline \begin{cases} \left | 2x+1 \right | &=-(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=+(2x-1) \end{cases}\\\hline \end{array}\\ &\textrm{Selanjutnya adalah} \end{aligned} \end{array}$
$.\qquad\begin{array}{|c|c|c|}\hline \begin{aligned}\: -\infty < x<& -\frac{1}{2}\: ,\\ 25-\left | 10x+5 \right |&\geq \left | 40x-20 \right |\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(-(2x+1))&\geq 4(-(2x-1))\\ 5+2x+1&\geq -8x+4\\ 10x&\geq -2\\ x&\geq -\frac{2}{10}\quad (\textbf{tm})\\ & \end{aligned}&\begin{aligned} \: -\frac{1}{2} \leq x<& \frac{1}{2}\: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1)&\geq 4(-(2x-1))\\ 5-2x-1&\geq -8x+4\\ 6x&\geq 0\\ x&\geq 0\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ 0\leq x< \frac{1}{2} \right \} \end{aligned} &\begin{aligned}\: \frac{1}{2}\leq x< &\infty \: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1))&\geq 4(2x-1)\\ 5-2x-1&\geq 8x-4\\ -10x&\geq -8\\ x&\leq \frac{8}{10}\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ \frac{1}{2}\leq x\leq \frac{4}{5} \right \} \end{aligned}\\\hline \end{array}$
$.\qquad\begin{aligned}&\\ &\textrm{Sehingga yang memenuhi}\: \textrm{adalah}:\\ &=\color{red}\left \{ 0\leq x\leq \displaystyle \frac{4}{5} \right \} \end{aligned}$.
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