PAS MATEMATIKA BLOG

Contoh Soal 1 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

$\begin{array}{ll}\\ 1.&\textrm{Diketahui pertidaksamaan}\: \: \displaystyle \frac{x+10}{x-9}\leq 0\\ &\textrm{dan diberikan beberapa nilai berikut}\\ &(\textrm{i})\quad x=-6\: \, \qquad\qquad (\textrm{iii})\quad x=-14\\ &(\textrm{ii})\, \, \, \: x=-10\qquad\quad\quad (\textrm{iv})\quad x=-18\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\textrm{di atas adalah ditunjukkan oleh}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&(\textrm{i})\: \: \textrm{dan} \: \: (\textrm{ii})\\ \textrm{b}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{c}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{d}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iv})\\ \textrm{e}.&(\textrm{iii})\: \: \textrm{dan}\: \: \: (\textrm{iv}) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x+10}{x-9}&\leq 0\\ \textrm{HP}=&\color{red}\left \{ x|-10\leq x< 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\leq x< 6\\ \textrm{b}.&2\leq x< 3\\ \color{red}\textrm{c}.&2< x< 3\: \: \textrm{atau}\: \: x>6\\ \textrm{d}.&x<3\: \: \textrm{atau}\: \: 3<x<6\\ \textrm{e}.&x<2\: \: \textrm{atau}\: \: x>3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\displaystyle \frac{6}{x-3}<\frac{8}{x-2}&\\ \displaystyle \frac{6}{x-3}-\frac{8}{x-2}&<0\\ \displaystyle \frac{6(x-2)-8(x-3)}{(x-3)(x-2)}&<0\\ \displaystyle \frac{6x-8x-12+24}{(x-2)(x-3)}&<0\\ \displaystyle \frac{-2x+12}{(x-2)(x-3)}&<0\\ \displaystyle \frac{2x-12}{(x-2)(x-3)}&>0\\ \displaystyle \frac{2(x-6)}{(x-2)(x-3)}&>0\\ \textrm{HP}=&\color{red}\left \{ x|2<x<3\: \: \textrm{atau}\: \: x>6,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{x^{2}-81}{x^{2}}\geq 0\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x\leq -9\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{b}.&-9\leq x< 0\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{c}.&-9\leq x< 0\: \: \textrm{atau}\: \: 0<x\leq 9\\ \textrm{d}.&-9< x\leq 9\\ \textrm{e}.&x\in \mathbb{R} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x^{2}-81}{x^{2}}&\geq 0\\ \displaystyle \frac{(x+9)(x-9)}{x^{2}}&\geq 0\\ \textrm{HP}=&\color{red}\left \{ x|x\leq -9\: \: \textrm{atau}\: \: x\geq 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}-4}{x+2}>0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x>2\\ \textrm{b}.&-2\leq x< 2\\ \textrm{c}.&x<-2\: \: \textrm{atau}\: \: x>2\\ \textrm{d}.&x<-2\: \: \textrm{atau}\: \: -2< x< 2\\ \textrm{e}.&x\geq -2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x^{2}-4}{x+2}&>0\\ \displaystyle \frac{(x+2)(x-2)}{(x+2)}&>0\\ (x-2)&>0\\ x&>2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}<0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|-9< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{b}.&\left \{ x|-6< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \}\\ \color{red}\textrm{d}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: \displaystyle \frac{5}{2}<x<5,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\left \{ x|x< -9\: \: \textrm{atau}\: \: -6< x< \displaystyle \frac{5}{2}\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}&<0\\ \displaystyle \frac{(x+6)(x-5)}{(x+9)(2x-5)}&<0\\ \textrm{Cukup jelas}& \end{aligned} \end{array}$

Pertidaksamaan Rasional dan Irasional Satu Variabel

A. Bentuk Umum Pertidaksamaan Rasional

$\begin{aligned}&\LARGE\textbf{Pertidak samaan Rasional}\\ &\\ &\begin{cases} \LARGE\textbf{A} & \begin{aligned}&\\ \begin{cases} \displaystyle \frac{f(x)}{g(x)} & <0 \\\\ \displaystyle \frac{f(x)}{g(x)} & \leq 0 \\\\ \displaystyle \frac{f(x)}{g(x)} & >0 \\\\ \displaystyle \frac{f(x)}{g(x)} & \geq 0 \end{cases}&\begin{matrix} \LARGE\textbf{misalnya} & \begin{cases} \displaystyle \frac{x-1}{x+3} & <0 \\\\ \displaystyle \frac{x-1}{x+3} & \leq 0 \\\\ \displaystyle \frac{x-1}{x+3} & >0 \\\\ \displaystyle \frac{x-1}{x+3} & \geq 0 \end{cases} \end{matrix}\\ & \end{aligned} \\ \LARGE\textbf{B} & \begin{aligned}&\\ \begin{cases} \displaystyle \frac{f(x)}{g(x)} & <0 \\\\ \displaystyle \frac{f(x)}{g(x)} & \leq 0 \\\\ \displaystyle \frac{f(x)}{g(x)} & >0 \\\\ \displaystyle \frac{f(x)}{g(x)} & \geq 0 \end{cases}&\begin{matrix} \LARGE\textbf{misalnya} & \begin{cases} \displaystyle \frac{x^{2}-4}{x+6} & <0 \\\\ \displaystyle \frac{x^{2}-4}{x+6} & \leq 0 \\\\ \displaystyle \frac{x^{2}-4}{x+6} & >0 \\\\ \displaystyle \frac{x^{2}-4}{x+6} & \geq 0 \end{cases} \end{matrix}\\ & \end{aligned} \\ \LARGE\textbf{C} & \begin{aligned}&\\ \begin{cases} \displaystyle \frac{f(x)}{g(x)} & <0 \\\\ \displaystyle \frac{f(x)}{g(x)} & \leq 0 \\\\ \displaystyle \frac{f(x)}{g(x)} & >0 \\\\ \displaystyle \frac{f(x)}{g(x)} & \geq 0 \end{cases}&\begin{matrix} \LARGE\textbf{misalnya} & \begin{cases} \displaystyle \frac{x^{2}+2x-3}{x^{2}-4} & <0 \\\\ \displaystyle \frac{x^{2}+2x-3}{x^{2}-4} & \leq 0 \\\\ \displaystyle \frac{x^{2}+2x-3}{x^{2}-4} & >0 \\\\ \displaystyle \frac{x^{2}+2x-3}{x^{2}-4} & \geq 0 \end{cases} \end{matrix}\\ & \end{aligned} \end{cases}\\ &\\ \end{aligned}$

B. Menyelesaikan Pertidaksamaan Rasional

Misal pada bentuk A di atas, maka penyelesaiannya adalah

$\begin{array}{|c|c|}\hline \begin{aligned}&\\ &\displaystyle \frac{x-1}{x+3}<0\\ & \end{aligned}&\begin{aligned}&\\ &\displaystyle \frac{x-1}{x+3}\leq 0\\ & \end{aligned} \\\hline \multicolumn{2}{|c|}{\Large\textbf{Wilayahnya}}\\\hline \begin{aligned} &\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{3-6} &+&&-&-&&+&&\\\hline &\multicolumn{2}{c}{-3}&&&\multicolumn{2}{l}{1}&& \end{array}\\ &\\ \textbf{HP}&=\left \{ x|-3<x<1,\: x\in \mathbb{R} \right \}\\ & \end{aligned}&\begin{aligned} &\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{3-6} &+&&-&-&&+&&\\\hline &\multicolumn{2}{c}{-3}&&&\multicolumn{2}{l}{\textcircled{1}}&& \end{array} \\ &\\ \textbf{HP}&=\left \{ x|-3<x\leq 1,\: x\in \mathbb{R} \right \}\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Untuk}&\: \textrm{bilangan yang dilingkari}\\ &\textrm{diartikan termasuk yang memenuhi}.\\ &\textrm{Jika tidak dilingkari maka tidak memenuhi}\\ &\end{aligned}}\\\hline \end{array}$

Jika nantinya berupa pertidaksamaan yang mengandung kuadrat, maka gunakan materi sebelumnya yang berkaitan dengan pertidaksamaan yang mengandung bentuk kuadrat.

Lihat di sini

C. Bentuk Umum Pertidaksamaan Irasional

Bentuk pertama

$\textbf{Bentuk Umum}\quad 1:\\\\ \: \: \sqrt{f(x)}\cdots A\: \begin{cases} \sqrt{f(x)}< A\\\\ \sqrt{f(x)}\leq A \\\\ \sqrt{f(x)}> A \\\\ \sqrt{f(x)}\geq A \end{cases}$

Bentuk kedua

$\textbf{Bentuk Umum}\quad 2:\\\\ \: \: \sqrt{f(x)}\cdots \sqrt{g(x)}\: \begin{cases} \sqrt{f(x)}< \sqrt{g(x)}\\\\ \sqrt{f(x)}\leq \sqrt{g(x)} \\\\ \sqrt{f(x)}> \sqrt{g(x)} \\\\ \sqrt{f(x)}\geq \sqrt{g(x)} \end{cases}$

D. Penyelesaian pada pertidaksamaan irasional

Kuadartakan masing-masing ruas
Dibawah tanda akar (numerus) haruslah $\geq 0 $
Himpunan penyelesaian berupa irisan dari penyelesaian yang didapatkan

Contoh Soal 3 Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

$\begin{array}{ll}\\ 11.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{5}{4x-3} \right |\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\leq x\leq \frac{3}{4}\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{b}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \displaystyle \frac{3}{4}< x\leq 2\\ \textrm{c}.&-\displaystyle \frac{1}{2}\leq x\leq 2,\: \: x\neq \frac{3}{4}\\ \textrm{d}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x>\frac{3}{4}\\ \color{red}\textrm{e}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x\geq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{5}{4x-3} \right |&\leq 1\\ -1\leq \displaystyle \frac{5}{4x-3}&\leq 1,\: \: \color{magenta}\textbf{jika dibalik}\\ -1\geq \displaystyle \frac{4x-3}{5}&\geq 1,\: \: \color{magenta}\textbf{bentuk ini tidak}\\ \color{magenta}\textbf{dibolehkan}&\: \color{magenta}\textbf{maka perlu diubah menjadi}\\ -1\geq \displaystyle \frac{4x-3}{5}\: \: \textrm{atau}&\: \: \displaystyle \frac{4x-3}{5}\geq 1,\: \: \color{black}\textrm{selanjutnya}\\ \bullet \quad \textrm{bagian}&\: 1\\ -1&\geq \displaystyle \frac{4x-3}{5}\Leftrightarrow \frac{4x-3}{5}\leq -1\\ 4x-3&\leq -5\\ 4x&\leq -2\\ x&\leq -\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{4x-3}{5}&\geq 1\\ 4x-3&\geq 5\\ 4x&\geq 8\\ x&\geq 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&(\textrm{UMPTN 95})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2}{2x-1} \right |> 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x> 2\\ \textrm{b}.&x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{c}.&x<-1\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{d}.&-1<x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{e}.&x<-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{semua opsi bukan jawaban}\\ &\textbf{Berikut pembahasannya}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{2}{2x-1} \right |&> 1\\ -1>\displaystyle \frac{2}{2x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2}{2x-1}>1,\: \color{magenta}\textbf{dibalik}\\ -1<\displaystyle \frac{2x-1}{2}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x-1}{2}<1\\ \bullet \quad \textrm{bagian}&\: 1\\ \displaystyle \frac{2x-1}{2}&>-1\\ 2x-1&>-2\\ 2x&>-1\\ x&>-\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{2x-1}{2}&<1\\ 2x-1&<2\\ 2x&<3\\ x&<\displaystyle \frac{3}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&(\textrm{UMPTN 00})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2x+7}{x-1} \right |\geq 1\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-2\leq x\leq 8\\ \textrm{b}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -2\\ \textrm{c}.&-8\leq x< 1\: \: \textrm{atau}\: \: x>1\\ \textrm{d}.&-2\leq x< 1\: \: \textrm{atau}\: \: 1< x\leq 8\\ \color{red}\textrm{e}.&x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{2x+7}{x-1} \right |&\geq 1\\ -1\geq \displaystyle \frac{2x+7}{x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x+7}{x-1}\geq 1\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{2x+7}{x-1}&\leq -1\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{2x+7}{x-1}&+1\leq 0\\ &\displaystyle \frac{2x+7+(x-1)}{x-1}\leq 0\\ \displaystyle \frac{3x+6}{x-1}&\leq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| -2\leq x< 1,\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{2x+7}{x-1}&\geq 1\\ \displaystyle \frac{2x+7}{x-1}&-1\geq 0\\ &\displaystyle \frac{2x+7-(x-1)}{x-1}\geq 0\\ \displaystyle \frac{x+8}{x-1}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>1,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x> 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{x-2}{x+3} \right |\leq 2\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-8\leq x< -3\\ \textrm{b}.&-8\leq x< -1\\ \textrm{c}.&-4\leq x< -3\\ \color{red}\textrm{d}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3}\\ \textrm{e}.&x\leq -4\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{x-2}{x+3} \right |&\leq 2\\ -2\leq \displaystyle \frac{x-2}{x+3}&\: \: \textrm{atau}\: \: \displaystyle \frac{x+2}{x+3}\leq 2\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{x-2}{x+3}&\geq -2\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{x-2}{x+3}&+2\geq 0\\ &\displaystyle \frac{x-2+2(x+3)}{x+3}\geq 0\\ \displaystyle \frac{3x+4}{x+3}&\geq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| x< -3\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{x-2}{x+3}&\leq 2\\ \displaystyle \frac{x-2}{x+3}&-2\leq 0\\ &\displaystyle \frac{x-2-2(x+3)}{x+3}\leq 0\\ \displaystyle \frac{-x-8}{x+3}&\leq 0,\: \: \color{magenta}\textbf{koefisien \textit{x} negatif}\\ \displaystyle \frac{x+8}{x+3}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>-3,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{2}{x+1}\leq \left | x \right |\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{b}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: 0\leq x\leq 1 \right \}\\ \textrm{c}.&\left \{ x|x\geq 1 \right \}\\ \color{red}\textrm{d}.&\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{e}.&\left \{ x|-1< x\leq 1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left | x \right |&\geq \displaystyle \frac{2}{x+1}\quad\quad\quad \color{black}\textrm{berakibat}\\ \displaystyle \frac{-2}{x+1}\geq x&\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{2}{x+1}\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ x\leq \displaystyle \frac{-2}{x+1}&\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ x+\displaystyle \frac{2}{x+1}&\leq 0\\ &\displaystyle \frac{x(x+1)+2}{x+1}\leq 0\\ \displaystyle \frac{x^{2}+x+2}{x+1}&\leq 0\Leftrightarrow \displaystyle \frac{\textrm{Definit positif}}{x+1}\leq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| x< -1,\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ x&\geq \displaystyle \frac{2}{x+3}\\ x-&\displaystyle \frac{2}{x+1}\geq 0\\ &\displaystyle \frac{x(x+1)-2}{x+1}\geq 0\\ &\displaystyle \frac{x^{2}+x-2}{x+1}\geq 0\\ &\displaystyle \frac{(x+2)(x-1)}{x+1}\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|-2\leq x< -1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

Lanjutan Materi Pertidaksamaan Berkaitan Materi Pertidaksamaan Kuadrat

Sebelumya

silahkan buka di sini

A. PERTIDAKSAMAAN KUADRAT

$\begin{array}{ll}\\ &\underline{\textbf{Bentuk Umum}}:\\\\ &\color{blue}\begin{cases} ax^{2}+bx+c< 0 \\ ax^{2}+bx+c\leq 0 \\ ax^{2}+bx+c > 0 \\ ax^{2}+bx+c \geq 0 \end{cases}\\ & \end{array}$

B. PENYELESAIAN PERTIDAKSAMAAN KUADRAT

$\begin{aligned}&\\ ax^{2}+bx+c\: \: ...\: \: 0&\\ \color{blue}\textrm{diubah menjadi}&\quad \: ax^{2}+bx+c=0\\ &\Leftrightarrow a\left ( x-x_{1} \right )\left ( x-x_{2} \right )=0\\ &\Leftrightarrow x=x_{1}\quad \textrm{atau}\quad x=x_{2}\\ &\end{aligned}$

$\begin{array}{|l|c|}\hline \textrm{Pertidaksamaan}&\textrm{Himpunan Penyelesaian dengan}\: \:x_{1}<x_{2} \\\hline ax^{2}+bx+c< 0&\left \{ x|x_{1}<x<x_{2},\: x\in \mathbb{R} \right \} \\\hline ax^{2}+bx+c\leq 0&\left \{ x|x_{1}\leq x\leq x_{2},\: x\in \mathbb{R} \right \} \\\hline ax^{2}+bx+c> 0&\left \{ x|x<x_{1}\: \: \textrm{atau}\: \: x>x_{2},\: x\in \mathbb{R} \right \} \\\hline ax^{2}+bx+c\geq 0&\left \{ x|x\leq x_{1}\: \: \textrm{atau}\: \: x\geq x_{2},\: x\in \mathbb{R} \right \} \\\hline \end{array}$

$\color{blue}\LARGE\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ &\textrm{Tentukan himpunan penyelesaian \textbf{PtKSV}}\\ &\textrm{(Pertidaksamaan Linear Satu Variabel) berikut ini!}\\ &\textrm{a}.\quad x^{2}-6x+8< 0\\ &\textrm{b}.\quad x^{2}-6x+8\leq 0\\ &\textrm{c}.\quad x^{2}-6x+8> 0\\ &\textrm{d}.\quad x^{2}-6x+8\geq 0\\\\ &\textrm{Jawab}:\\ \end{array}$

$\begin{aligned}&\\ x^{2}-6x+8\: \: ...\: \: 0&\\ \color{blue}\textrm{diubah menjadi}&\quad \: x^{2}-6x+8=0\\ &\Leftrightarrow 1\left ( x-2 \right )\left ( x-4 \right )=0\\ &\Leftrightarrow x=2\quad \textrm{atau}\quad x=4\\ &\end{aligned}$

$\begin{array}{ll}\\ &\begin{array}{|l|c|c|}\hline \textrm{Pertidaksamaan}&\textrm{HP dengan}\: \:\left (2<4 \right )&\textrm{Selang/Interval}\\\hline x^{2}-6x+8< 0&\left \{ x|2<x<4,\: x\in \mathbb{R} \right \}&\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{3-6} &+&&-&-&&+&+&\\\hline &\multicolumn{2}{r}{2}&&&\multicolumn{2}{l}{4}&& \end{array} \\\hline x^{2}-6x+8\leq 0&\left \{ x|2\leq x\leq 4,\: x\in \mathbb{R} \right \}&\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{3-6} &+&&-&-&&+&+&\\\hline &\multicolumn{2}{r}{\textcircled{2}}&&&\multicolumn{2}{l}{\textcircled{4}}&& \end{array} \\\hline x^{2}-6x+8> 0&\left \{ x|x<2\: \: \textrm{atau}\: \: x>4,\: x\in \mathbb{R} \right \}&\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{1-2}\cline{7-9} &+&&-&-&&+&+&\\\hline &\multicolumn{2}{r}{2}&&&\multicolumn{2}{l}{4}&& \end{array} \\\hline x^{2}-6x+8\geq 0&\left \{ x|x\leq 2\: \: \textrm{atau}\: \: x\geq 4,\: x\in \mathbb{R} \right \}&\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{1-2}\cline{7-9} &+&&-&-&&+&+&\\\hline &\multicolumn{2}{r}{\textcircled{2}}&&&\multicolumn{2}{l}{\textcircled{4}}&& \end{array} \\\hline \multicolumn{3}{|c|}{\begin{aligned}\textbf{Catatan: }&\textrm{Angka yang dilingkari termasuk}\\ &\textrm{himpunan penyelesaian} \end{aligned}}\\\hline \end{array} \end{array}$

Contoh Soal 2 Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

$\begin{array}{ll}\\ 6.&\textrm{Penyelesaian dari pertidaksamaan}\\ &\left | \displaystyle \frac{3-2x}{-5} \right |>5\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-11\: \: \textrm{atau}\: x>14\\ \textrm{b}.&x<-14\: \: \textrm{atau}\: x>11\\ \textrm{c}.&11<x<14\\ \textrm{d}.&-14<x<-11\\ \textrm{e}.&x>14 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{3-2x}{-5} \right |&>5\\ \displaystyle \frac{3-2x}{-5}<&-5\: \: \textrm{atau}\: \: \displaystyle \frac{3-2x}{-5}>5\\ \displaystyle \frac{2x-3}{5}>&5\: \: \textrm{atau}\: \: \displaystyle \frac{2x-3}{5}<-5\\ 2x-3>&25\: \: \textrm{atau}\: \: 2x-3<-25\\ 2x>25&+3\: \: \textrm{atau}\: \: 2x<-25+3\\ x>14&\: \: \textrm{atau}\: \: x<-11,\\ &\textrm{dapat juga dituliskan}\\ x<-&11\: \: \textrm{atau}\: \: x>14 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 2-2\left | x+1 \right | \right |>4\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-4\: \: \textrm{atau}\: x>2\\ \textrm{b}.&x<-3\: \: \textrm{atau}\: x>1\\ \textrm{c}.&x<-2\: \: \textrm{atau}\: x>0\\ \textrm{d}.&x<-1\: \: \textrm{atau}\: x>3\\ \textrm{e}.&x<0\: \: \textrm{atau}\: x>4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\left | 2-2\left | x+1 \right | \right |>4&\\ 2-2\left | x+1 \right |<-4&\: \: \textrm{atau}\: \: 2-2\left | x+1 \right |>4\\ -2\left | x+1 \right |<-6\: \: &\textrm{atau}\: \: -2\left | x+1 \right |>2\\ \left | x+1 \right |>3\: \: &\textrm{atau}\: \: \left | x+1 \right |<-1\\ \left\{\begin{matrix} (x+1)<-3\\ (x+1)>3 \end{matrix}\right.\: \: &\textrm{atau}\: \: \left\{\begin{matrix} \left | x+1 \right |<-1\\ \color{red}\textbf{tak mungkin} \end{matrix}\right.\\ \textrm{Selanjutnya}&\: \textrm{akan didapatkan}\\ x<-4\: \: &\textrm{atau}\: \: x>2 \end{aligned} \end{array}$

$\begin{array}{l}\\ 8.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 3-\left | x \right | \right |<10\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-14\: \: \textrm{atau}\: x>12\\ \textrm{b}.&x<-13\: \: \textrm{atau}\: x>13\\ \textrm{c}.&x<-12\: \: \textrm{atau}\: x>10\\ \textrm{d}.&0<x<10\\ \color{red}\textrm{e}.&-13<0<13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\left | 3-\left | x \right | \right |&< 10\\ -10< 3-&\left | x \right |< 10\\ -13< -&\left | x \right |< 7\\ -7< &\left | x \right |\leq 13,\: \: (\textrm{ingat harga}\: \: \left | x \right |\geq 0)\\ 0\leq &\left | x \right |< 13\\ &\textrm{selanjutnya},\\ &\left | x \right |< 13\\ -13< &\: x< 13\\ \textrm{HP}=&\color{red}\left \{ x|\: -13< x< 13,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{l}\\ 9.&\textrm{(UM UGM 05)}\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x^{2}-3 \right |<2x\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<3\\ \textrm{b}.&-3<x<1\\ \color{red}\textrm{c}.&1<x<3\\ \textrm{d}.&-3<x<-1\: \: \textrm{atau}\: \: 1<x<3\\ \textrm{e}.&x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\left | x^{2}-3 \right |&<2x\\ -2x<\left ( x^{2}-3 \right )&<2x\\ \textrm{dipartisi men}&\textrm{jadi dua bagian}\\ \bullet \quad\textrm{pertama}\qquad&\\ (x^{2}-3)&>-2x\\ x^{2}+2x-3&>0\\ (x+3)(x-1)&>0\\ x<-3\: \: \textrm{atau}&\: \: x>1\\ \bullet \quad \textrm{kedua}\qquad\quad&\\ \left ( x^{2}-3 \right )&<2x\\ x^{2}-2x-3&<0\\ (x-3)(x+1)&<0\\ -1<x<3&\\ \textrm{ambil yang}&\: \textrm{memenuhi keduanya}\\ \textrm{berupa iris}&\textrm{an}\\ \textrm{HP}=&\color{red}\left \{ 1<x<3,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&(\textrm{SPMB 05})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x-2 \right |^{2}<4\left | x-2 \right |+12\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x\in \mathbb{R}|2\leq x\leq 8 \right \}\\ \textrm{b}.&\left \{ x\in \mathbb{R}|4<x< 8 \right \}\\ \color{red}\textrm{c}.&\left \{ x\in \mathbb{R}|-4<x< 8 \right \}\\ \textrm{d}.&\left \{ x\in \mathbb{R}|-2<x<4 \right \}\\ \textrm{e}.&\left \{ x\in \mathbb{R}|2<x<4 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{misalkan}\: \: p&=\left | x-2 \right |,\: \: \textrm{selanjutnya}\\ \left | x-2 \right |^{2}<&\, 4\left | x-2 \right |+12\\ p^{2}<&\, 4p+12\\ p^{2}-&4p-12<0\\ (p-6)&(p+2)<0\\ -2<p&<6,\: \: \color{magenta}\textrm{atau jika dikembalikan}\\ -2<&\left | x-2 \right |<6,\: \: \color{black}\textrm{ingat, nilanya tidak negatif}\\ 0\leq &\left | x-2 \right |<6\\ -6<&\: x-2<6\\ -4<&\: x<8\\ \textrm{HP}=&\color{red}\left \{ -4<x<8,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

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

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

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

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

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

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

Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

$\LARGE\textrm{A. Sifat-Sifat yang Berlaku}$

Untuk a > 0, maka

berlaku sifat-sifat sebagai berikut:

$\begin{cases} \left | x \right |< a & \Leftrightarrow -a< x< a \\ \left | x \right |\leq a & \Leftrightarrow -a\leq x\leq a \\ \left | x \right |> a & \Leftrightarrow x< -a\: \: \: \textrm{atau}\: \: \: a> a \\ \left | x \right |\geq a & \Leftrightarrow x\leq -a\: \: \: \textrm{atau}\: \: \: a\geq a\\\\ \qquad\textrm{dan}&\textrm{perlu diingat}\\ \qquad\left | x \right |&=\sqrt{x^{2}} \end{cases}$

$\LARGE\textrm{B. Penyelesaian Pertidaksamaan}$

$\color{blue}\LARGE\textrm{a. bentuk pertama}$

$\left | f(x) \right |\begin{cases} < a & \Rightarrow -a\leq f(x)\leq a\\ \leq a& \Rightarrow -a\leq f(x)\leq a\\ > a & \Rightarrow f(x)< -a\: \: \textrm{atau}\: \: f(x)> a\\ \geq a & \Rightarrow f(x)\leq -a\: \: \textrm{atau}\: \: f(x)\geq a \end{cases}$

$\color{magenta}\LARGE\fbox{CONTOH SOAL}$

$\begin{array}{l}\\ \textrm{Tentukanlah himpunan penyelesaian dari}\\ (1).\quad \left | 3x-7 \right |\leq 11\\ (2).\quad \left | 3x-7 \right |> 11\\\\ \textrm{Jawab}:\\\end{array}$

$\begin{aligned}(1).\qquad\qquad\qquad\left | 3x-7 \right |&\leq 11\\ -11\leq 3x-7&\leq 11\\ -11+(7)\leq 3x-7+(7)&\leq 11+(7)\\ -4\leq 3x&\leq 18\qquad (\textrm{dibagi})\: \: 3\\ -\displaystyle \frac{4}{3}\leq x&\leq 6\\ \textrm{HP}=&\color{red}\left \{ x|-\displaystyle \frac{4}{3}\leq x\leq 6,\: x\in \mathbb{R} \right \} \end{aligned}$

$\begin{aligned}(2).\quad\quad\left | 3x-7 \right |&> 11\\ (3x-7)<-11&\quad \textrm{atau}\quad (3x-7)>11\\ 3x<-11+7&\: \: \textrm{atau}\: \: 3x>11+7\\ x<-\displaystyle \frac{4}{3}\: \: \textrm{atau}&\: \: x>6\\ \textrm{HP}=&\color{red}\left \{ x|x<-\displaystyle \frac{4}{3}\: \: \textrm{atau}\: \: x>6,x\in \mathbb{R} \right \} \end{aligned}$

$\color{blue}\LARGE\textrm{b. bentuk kedua}$

$\left | f(x) \right |\begin{cases} < \left | g(x) \right | & \\ \leq \left | g(x) \right |& \\ > \left | g(x) \right | & \\ \geq \left | g(x) \right | & \end{cases}\Rightarrow \textrm{dikuadratkan kedua ruas}$

Perlu diingat di sini bahwa bentuk selanjutnya jika berupa pertidaksamaan kuadrat, maka langkah selanjutnya perlu diperhatikan langkah-langkah berikut:

$\color{magenta}\begin{aligned}ax^{2}+bx+c&\: \: \begin{cases} < \\ \leq \\ > \\ \geq \end{cases} 0\\ \textrm{dengan}\: &a, \: b,\: c\: \in \mathbb{R},\: a\neq 0 \end{aligned}$

dan juga

$\textrm{Akar-akar}\: \begin{cases} x_{1} & \text{ dan }\: \: x_{2}\: \: \textrm{dengan}\: \: x_{1}\neq x_{2} \\ & \text{ serta} \\ x_{1} &\leq \: \: x_{2} \end{cases}$

maka dalam menentukan himpunan penyelesaiannya adalah sebagai berikut:

$\begin{cases} ax^{2}+bx+c< 0, &\textrm{HP}=\left \{ x|x_{1}< x< x_{2},\: x\in \mathbb{R} \right \} \\ ax^{2}+bx+c\leq 0, &\textrm{HP}=\left \{ x|x_{1}\leq x\leq x_{2},\: x\in \mathbb{R} \right \} \\ ax^{2}+bx+c> 0, &\textrm{HP}=\left \{ x|x< x_{1}\: \textrm{atau}\: x> x_{2},\: x\in \mathbb{R} \right \} \\ ax^{2}+bx+c\geq 0, &\textrm{HP}=\left \{ x|x\leq x_{1}\: \textrm{atau}\: x\geq x_{2},\: x\in \mathbb{R} \right \} \end{cases}$

$\color{magenta}\LARGE\fbox{CONTOH SOAL}$

$\begin{array}{l}\\ \textrm{Tentukanlah himpunan penyelesaian dari}\\ (1).\quad \left | 3x-7 \right |\leq \left | x-1 \right |\\ (2).\quad \left | 3x-7 \right |> \left | x-1 \right |\\\\ \textrm{Jawab}:\\\end{array}$

$\begin{aligned}(1).\: \left | 3x-7 \right |\leq \left | x-1 \right |&\\ \left | 3x-7 \right |^{2}\leq \left | x-1 \right |^{2}&\\ (3x-7)^{2}-(x-1)^{2}&\leq 0\\ (3x-7+x-1)&(3x-7-(x-1))\leq 0\\ (4x-8)(2x-6)&\leq 0\\ (x-2)(x-3)&\leq 0\\ \textrm{HP}=&\color{red}\left \{ x|2\leq x\leq 3,\: x\in \mathbb{R} \right \} \end{aligned}$

$\begin{aligned}(2).\quad \left | 3x-7 \right |&> \left | x-1 \right |\\ \left | 3x-7 \right |^{2}&> \left | x-1 \right |^{2}\\ (3x-7)^{2}-&(x-1)^{2}> 0\\ (3x-7+&x-1)(3x-7-(x-1))> 0\\ (4x-8)&(2x-6)> 0\\ (x-2)&(x-3)> 0\\ \textrm{HP}=&\color{red}\left \{ x|x<2\: \: \textrm{atau}\: \: x>3,\: x\in \mathbb{R} \right \} \end{aligned}$

$\color{blue}\LARGE\textrm{c. bentuk ketiga}$

Penyelesaian jenis ini mirip poin yang pertama, yaitu:

$\begin{array}{l}\\ \left | f(x) \right |\begin{cases} \left\{\begin{matrix} < g(x)\\ \leq g(x) \end{matrix}\right. & \text{ maka }\left\{\begin{matrix} -g(x)<f(x)<g(x)\\ -g(x)\leq f(x)\leq g(x) \end{matrix}\right. \\\\ \left\{\begin{matrix} > g(x)\\ \geq g(x) \end{matrix}\right. & \text{ maka } \left\{\begin{matrix} \left\{\begin{matrix} f(x)>g(x)\: \: \: \\ f(x)<-g(x) \end{matrix}\right.\\\\ \left\{\begin{matrix} f(x)\geq g(x)\: \: \: \\ f(x)\leq -g(x) \end{matrix}\right. \end{matrix}\right. \end{cases} \end{array}$

$\color{magenta}\LARGE\fbox{CONTOH SOAL}$

$\begin{array}{l}\\ \textrm{Tentukanlah himpunan penyelesaian dari}\\ (1).\quad \left | 3x-7 \right |\leq (x-1)\\ (2).\quad \left | 3x-7 \right |> (x-1)\\\\ \textrm{Jawab}:\\\end{array}$

$\begin{aligned}(1).-g(x)\leq f(x)&\leq g(x)\\ \: (\textrm{a}).\: \: \textrm{untuk}\: \: f(x)&\geq -g(x)\\ \: \qquad 3x-7&\geq -(x-1)\\ 4x&\geq 8\\ x&\geq 2\\ (\textrm{b}).\: \: \textrm{untuk}\: \: f(x)&\leq g(x)\\ 3x-7&\leq x-1\\ 2x&\leq 6\\ x&\leq 3\\ \textrm{HP}=&\color{red}\left \{ x|2\leq x\leq 3,\: x\in \mathbb{R} \right \} \end{aligned}$

$\begin{aligned}(2).\: f(x)<-g(x)&\: \: \textrm{atau}\: \: f(x)> g(x)\\ \: (\textrm{a}).\: \: \textrm{untuk}\: \: f(x)&< -g(x)\\ \: \qquad 3x-7&<-(x-1)\\ 4x&< 8\\ x&< 2\\ (\textrm{b}).\: \: \textrm{untuk}\: \: f(x)&> g(x)\\ 3x-7&> x-1\\ 2x&> 6\\ x&> 3\\ \textrm{HP}=&\color{red}\left \{ x|x< 2\: \: \textrm{atau}\: \: x>3,\: x\in \mathbb{R} \right \} \end{aligned}$

Daftar Pustaka

Kanginan, M. 2016. Matematika untuk Siswa SMA-MA/SMK-MAK Kelas X (Wajib). Bandung: Srikandi Empat Widya Utama.
Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata PElajaran Wajib. Solo: PT Tiga Serangkai Pustaka Mandiri.

Contoh Soal Lanjutan 4 Nilai Mutlak

$\begin{array}{ll}\\ 21.&\textrm{Jumlah akar-akar persamaan}\\ &x^{2}+\left | x \right |-6=0,\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \color{red}\textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}x^{2}+\left | x \right |&-6=0\\ \left | x \right |^{2}+\left | x \right |&-6=0\\ \left ( \left | x \right |+3 \right )&\left ( \left | x \right |-2 \right )=0\\ \left | x \right |=-3&\: \: \textrm{atau}\: \: \left | x \right |=2\\ \textrm{tidak meme}&\textrm{nuhi (-3) atau}\: \: x=\pm 2\\ &\begin{cases} x_{1} & =2 \\ x_{2} & =-2 \end{cases}\\ x_{1}+x_{2}&=2+(-2)=0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Penyelesaian untuk}\: \: -2\left | x-5 \right |=8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 0 \right \}\\ \color{red}\textrm{b}.&\left \{ \: \: \right \}\\ \textrm{c}.&\left \{ 4,5 \right \}\\ \textrm{d}.&\left \{ 1,9 \right \}\\ \textrm{e}.&\left \{ -1,9 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}-2\left | x-5 \right |&=8\\ \left | x-5 \right |&=-4\\ \textrm{Karena pe}&\textrm{rsamaan bernilai negatif},\\ \textrm{maka tida}&\textrm{k ada nilai}\: \: x\: \: \textrm{yang memenuhi}\\ \therefore \: \textbf{HP}&=\left \{ \: \: \right \} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \left |\displaystyle \frac{x-5}{2x+1} \right |=2\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{5}{3}\: \: \textrm{atau}\: \: \frac{3}{5}\\ \textrm{b}.&\displaystyle \frac{5}{3}\: \: \textrm{atau}\: \: -\frac{3}{5}\\ \color{red}\textrm{c}.&-\displaystyle \frac{7}{3}\: \: \textrm{atau}\: \: \frac{3}{5}\\ \textrm{d}.&\displaystyle \frac{7}{3}\: \: \textrm{atau}\: \: -\frac{3}{5}\\ \textrm{e}.&-\displaystyle \frac{7}{3}\: \: \textrm{atau}\: \: \frac{3}{7}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\left |\displaystyle \frac{x-5}{2x+1} \right |&=2\\ \displaystyle \frac{x-5}{2x+1} &=\pm 2\\ \color{black}\textrm{untuk}&\: \: x=2\\ \displaystyle \frac{x-5}{2x+1} &=2\\ x-5&=2(2x+1)\\ x-4x&=2+5\\ -3x&=7\\ x&=\displaystyle \frac{7}{-3}=-\frac{7}{3}\\ \color{black}\textrm{untuk}&\: \: x=-2\\ x-5&=-2(2x+1)\\ x+4x&=-2+5\\ 5x&=3\\ x&=\displaystyle \frac{3}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \left |\left | 5x-4 \right |-3 \right |=2\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-3\: \: \textrm{atau}\: \: -6\\ \color{red}\textrm{b}.&-3\: \: \textrm{atau}\: \: 6\\ \textrm{c}.&3\: \: \textrm{atau}\: \: -6\\ \textrm{d}.&3\: \: \textrm{atau}\: \: 6\\ \textrm{e}.&6\: \: \textrm{atau}\: \: 9\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\left |\left | 2x-3 \right |-4 \right |&=5\\ \left | 2x-3 \right |-4&=\pm 5\\ \left | 2x-3 \right |&=\pm 5+4\: \: \color{black}\textrm{maka},\\ \left | 2x-3 \right |&=5+4=9\: \: (\textrm{memenuhi})\\ \left | 2x-3 \right |&=-5+4=-1\\ &\color{magenta}(\textbf{tidak memenuhi})\\ \color{black}\textrm{Selanjutnya}&,\\ (2x-3)&=\pm 9\\ 2x&=\pm 9+3\\ \color{black}\textrm{untuk}&\: \: x=9\\ 2x&=9+3\\ x&=\displaystyle \frac{12}{2}=6\\ \color{black}\textrm{untuk}&\: \: x=-9\\ 2x&=-9+3\\ x&=\displaystyle \frac{-6}{2}=-3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\textrm{persamaan}\: \: \left | x+1 \right |-2\left | x-3 \right | =5\left | x-4 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\frac{1}{6}\: \: \: \textrm{atau}\: \: \: 1\frac{2}{3}\\\\ \textrm{b}.&\displaystyle 1\frac{5}{6}\: \: \: \textrm{atau}\: \: \: 2\frac{2}{5}\\\\ \textrm{c}.&\displaystyle 1\frac{5}{6}\: \: \: \textrm{atau}\: \: \: 2\frac{2}{3}\\\\ \textrm{d}.&\displaystyle 2\frac{3}{4}\: \: \: \textrm{atau}\: \: \: 4\frac{2}{3}\\\\ \color{red}\textrm{e}.&\displaystyle 3\frac{1}{4}\: \: \: \textrm{atau}\: \: \: 4\frac{1}{2} \end{array}\\\\ &\textrm{Jawab}:\qquad \textrm{e}\\ &\color{blue}\begin{aligned}\left | x+1 \right |-&2\left | x-3 \right | =5\left | x-4 \right |\\ \textrm{Perhatik}&\textrm{an untuk batas sesuai definisi, maka}\\ x=-1&,\: x=3,\: \: x=4\\ \color{black}\textrm{saat}\: \: \: x&\geq 4\\ \left | x+1 \right |&=x+1,\: \: \left | x-3 \right |=x-3,\: \: \left | x-4 \right |=x-4\\ (x+1)-&2(x-3)=5(x-4)\\ x-2x-&5x=-1-6-20\\ -6x&=-27\\ x&=\displaystyle \frac{27}{6}=4\frac{3}{6}\: \: \color{magenta}(\textrm{memenuhi}) \\ \color{black}\textrm{saat}\: \: \: 3&\leq x< 4\\ \left | x+1 \right |&=x+1,\: \: \left | x-3 \right |=x-3,\: \: \left | x-4 \right |=4-x\\ (x+1)-&2(x-3)=5(4-x)\\ x-2x+&5x=-1-6+20\\ 4x&=13\\ x&=\displaystyle \frac{13}{4}=3\frac{1}{4}\: \: \color{magenta}(\textrm{memenuhi})\\ \color{black}\textrm{saat}\: \: \: -&1\leq x< 3\\ \left | x+1 \right |&=x+1,\: \: \left | x-3 \right |=3-x,\: \: \left | x-4 \right |=4-x\\ (x+1)-&2(3-x)=5(4-x)\\ x+2x+&5x=-1+6+20\\ 8x&=25\\ x&=\displaystyle \frac{25}{8}=3\frac{1}{8}\: \: \color{magenta}(\textrm{tidak memenuhi})\\ \color{black}\textrm{saat}\: \: \: -&1>x\\ \left | x+1 \right |&=-x-1,\: \: \left | x-3 \right |=3-x,\: \: \left | x-4 \right |=4-x\\ (-x-1)&-2(3-x)=5(4-x)\\ -x+2x&+5x=1+6+20\\ 6x&=27\\ x&=\displaystyle \frac{27}{6}=4\frac{3}{6}\: \: \color{magenta}(\textrm{tidak memenuhi}) \end{aligned} \end{array}$

Sumber Referensi

Kanginan, M. 2016. Matematika untuk Siswa SMA-MA/SMK-MAK Kelas X. Bandung: Srikandi Empat Widya Utama.
Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.

Contoh Soal Lanjutan 3 Nilai Mutlak

$\begin{array}{l}\\ 16.&\textrm{Penyelesaian dari}\: \: \left | 3x-(4x-7) \right |=6\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ -1,-13 \right \}\\ \textrm{b}.&\left \{ -1,13 \right \}\\ \color{red}\textrm{c}.&\left \{ 1,13 \right \}\\ \textrm{d}.&\left \{ -13,1 \right \}\\ \textrm{e}.&\left \{ 13 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\left | 3x-(4x-7) \right |&=6\\ \left | 3x-4x+7 \right |&=6\\ \left | -x+7 \right |&=6\\ (-x+7)&=\pm 6\\ -x&=\pm 6-7\\ \textrm{dikalikan}\: \: & \textrm{dengan}\: \: \color{black}-1\\ x&=\pm 6+7\\ x_{1}&=+6+7=+13\\ x_{2}&=-6+7=+1 \end{aligned} \end{array}$

$\begin{array}{l}\\ 17.&\textrm{Penyelesaian}\: \: \left | x-1 \right |=2x+1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ -2 \right \}\\ \textrm{b}.&\left \{ -2,0 \right \}\\ \textrm{c}.&\left \{ -1 \right \}\\ \textrm{d}.&\left \{ \: \: \, \right \}\\ \color{red}\textrm{e}.&\left \{ 0 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\left | x-1 \right |&=2x+1\\ (x-1)&=\pm (2x+1)\\ \textrm{untuk}\: &\begin{cases} x\geq 1 & \text{ maka } (x-1)=2x+1 \\ x< 1 & \text{ maka } (x-1)=-(2x+1) \end{cases}\\ &\begin{array}{|c|c|}\hline x\geq 1&x< 1\\\hline \begin{aligned}x-1&=2x+1\\ x-2x&=1+1\\ -x&=2\\ x&=-2\\ & \end{aligned}&\begin{aligned}x-1&=-(2x+1)\\ x-1&=-2x-1\\ x+2x&=-1+1\\ 3x&=0\\ x&=0 \end{aligned}\\\hline \textrm{Tidak memenuhi}&\textrm{memenuhi}\\\hline \end{array} \end{aligned} \end{array}$

$\begin{array}{l}\\ 18.&\textrm{Penyelesaian}\: \: \left | 3a+1 \right |=2a+9\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ -2 \right \}\\ \textrm{b}.&\left \{ 8 \right \}\\ \color{red}\textrm{c}.&\left \{ -2,8 \right \}\\ \textrm{d}.&\left \{ \: \: \, \right \}\\ \textrm{e}.&\textrm{Semua bilangan riil} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\left | 3a+1 \right |&=2a+9\\ (3a+1)&=\pm (2a+9)\\ \textrm{untuk}\: &\begin{cases} a\geq -\displaystyle \frac{1}{3} & \text{ maka } (3a+1)=2a+9 \\\\ a< -\displaystyle \frac{1}{3} & \text{ maka } (3a+1)=-(2a+9) \end{cases}\\ &\begin{array}{|c|c|}\hline a\geq \displaystyle -\frac{1}{3}&a< -\displaystyle \frac{1}{3}\\\hline \begin{aligned}3a+1&=2a+9\\ 3a-2a&=9-1\\ a&=8\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}3a+1&=-(2a+9)\\ 3a+1&=-2a-9\\ 3a+2a&=-9-1\\ 5a&=-10\\ a&=\displaystyle \frac{-10}{5}\\ a&=-2 \end{aligned}\\\hline \textrm{memenuhi}&\textrm{memenuhi}\\\hline \end{array} \end{aligned} \end{array}$

$\begin{array}{l}\\ 19.&\textrm{Penyelesaian dari}\: \: \left | 3x+2 \right |=4x+5\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-3\\ \color{red}\textrm{b}.&-1\\ \textrm{c}.&-3\: \: \textrm{dan}\: \: -1\\ \textrm{d}.&\left \{ \: \: \right \} \\ \textrm{e}.&0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\color{blue}\begin{aligned}\left | 3x+2 \right |&=4x+5\\ (3x+2)&=\pm (4x+5)\\ \textrm{untuk}\: &\begin{cases} x\geq -\displaystyle \frac{2}{3} & \text{ maka } 3x+2=4x+5 \\\\ x< -\displaystyle \frac{2}{3} & \text{ maka } 3x+2=-(4x+5) \end{cases}\\ &\begin{array}{|c|c|}\hline x\geq -\displaystyle \frac{2}{3}&x< -\displaystyle \frac{2}{3}\\\hline \begin{aligned}3x+2&=4x+5\\ 3x-4x&=5-2\\ -&=3\\ x&=-3\\ &\\ & \end{aligned}&\begin{aligned}3x+2&=-(4x+5)\\ 3x+2&=-4x-5\\ 3x+4x&=-5-2\\ 7x&=-7\\ x&=\displaystyle \frac{-7}{7}\\ x&=-1 \end{aligned}\\\hline \end{array} \end{aligned} \end{aligned} \end{array}$

$\begin{array}{l}\\ 20.&\textrm{Jika}\: \: \left | -3x \right |+4y^{-1}=6z+4x,\\ &\textrm{maka nilai}\: \: x\: \: \textrm{dinyatakan dalam}\: \: \: y\: \: \textrm{dan}\: \: z\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{y}{4}-6z\\ \color{red}\textrm{b}.&\displaystyle \frac{4}{y}-6z\\ \textrm{c}.&4-\displaystyle \frac{6}{y}\\ \textrm{d}.&\displaystyle \frac{4}{7y}+\frac{6z}{7}\\ \textrm{e}.&\displaystyle \frac{7}{4y}+6z \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\left | -3x \right |&+4y^{-1}=6z+4x\\ \left | -3x \right |&=6z+4x-4y^{-1}\\ -3x&=\pm (6z+4x-4y^{-1})\\ \textrm{untuk}&:\: x\geq 0\\ -3x&=6z+4x-4y^{-1}\\ -3x-4x&=6z-4y^{-1}\\ -7x&=6z-4y^{-1}\\ x&=\displaystyle \frac{6z-4y^{-1}}{-7}\\ &=\frac{4}{7y}-\frac{6z}{7}\\ untuk&:\: x<0\\ -3x&=-(6z+4x-4y^{-1})\\ -3x&=-6z-4x+4y^{-1}\\ -3x+4x&=-6z+4y^{-1}\\ x&=4y^{-1}-6z\\ &=\displaystyle \frac{4}{y}-6z \end{aligned} \end{array}$

Contoh Soal Lanjutan 2 Nilai Mutlak

$\begin{array}{l}\\ 11.&\textrm{Nilai}\: \: p\: \: \textrm{yang memenuhi}\: \: 10-4\left | 4-5p \right |=26\: \: \textrm{adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&2\: \: \textrm{atau}\: \: 1\displaystyle \frac{2}{3}\\ \textrm{b}.&1\: \: \textrm{atau}\: \: -\displaystyle \frac{3}{5}\\\\ \textrm{c}.&2\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\\ \textrm{d}.&-2\displaystyle \frac{3}{4}\: \: \textrm{atau}\: \: 1\\ \color{red}\textrm{e}.&-1\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}10-4\left | 4-5p \right |&=-26\\ -4\left | 4-5p \right |&=-36\\ \left | 4-5p \right |&=9\\ (4-5p)&=\pm 9\\ -5p&=-4\pm 9\\ p&=\displaystyle \frac{-4\pm 9}{-5}\\ p&=\begin{cases} \displaystyle \frac{-4+9}{-5} & =-1 \\ \textrm{atau} & \\ \displaystyle \frac{-4-9}{-5} & = \displaystyle \frac{13}{5}=2\frac{3}{5} \end{cases} \end{aligned} \end{array}$

$\begin{array}{l}\\ 12.&\textrm{Jika}\: \: 3<x<5\: \: \textrm{maka penyelesaian untuk}\\ &\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{lllllll}\\ \textrm{a}.&2x-2\\ \textrm{b}.&2\\ \textrm{c}.&8-2x\\ \textrm{d}.&-2\\ \color{red}\textrm{e}.&2x-8\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\sqrt{x^{2}-6x+9}&-\sqrt{x^{2}-10x+25}\\ &\sqrt{(x-3)^{2}}-\sqrt{(x-5)^{2}}\\ &=\left | x-3 \right |-\left | x-5 \right |\\ &\textrm{ingat bahwa saat}\\ \: \: 3<x<5\: \: &\textrm{maka}\\ &\: \begin{cases} \left | x-3 \right |=(x-3) \\ \left | x-5 \right |=-(x-5) \end{cases},\\ &\textrm{sehingga}\\ &=\left | x-3 \right |-\left | x-5 \right |\\ &=(x-3)-\left ( -(x-5) \right )\\ &=x-3+x-5\\ &=2x-8 \end{aligned} \end{array}$

$\begin{array}{l}\\ 13.&\textrm{Jika}\: \: 1<x<5\: \: \textrm{maka penyelesaian untuk}\\ &\sqrt{x^{2}-2x+1}+\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{lllllll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \color{red}\textrm{c}.&4\\ \textrm{d}.&5\\ \textrm{e}.&6\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\sqrt{x^{2}-2x+1}&+\sqrt{x^{2}-10x+25}\\ &=\sqrt{(x-1)^{2}}+\sqrt{(x-5)^{2}}\\ &=\left | x-1 \right |+\left | x-5 \right |\\ &\textrm{ingat bahwa saat}\\ 1<x<5\: \: &\textrm{maka}\: \begin{cases} \left | x-1 \right |=(x-1) \\ \left | x-5 \right |=-(x-5) \end{cases},\\ & \textrm{sehingga}\\ &=\left | x-1 \right |+\left | x-5 \right |\\ &=(x-1)+\left ( -(x-5) \right )\\ &=x-1+5-x\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Perhatikanlah ilustrasi grafik di bawah ini}\\ &\begin{aligned}\end{aligned}\end{array}$

$\begin{array}{l}\\ .\: \: \: \: \: \: &\textrm{Persamaan yang memenuhi rumus tersebut adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=\left | -x-2 \right |\\ \textrm{b}.&y=-2x-4\\ \textrm{c}.&y=-\left | 2x-4 \right |\\ \textrm{d}.&y=\left | -2x-4 \right |\\ \color{red}\textrm{e}.&y=\left | -2x+4 \right | \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\textrm{Dengan cara substitusi langsung}\\ & \textrm{kita akan mendapatkan}\\ &\bullet \quad \textrm{untuk}\: \: x=4\: \: \textrm{menyebabkan nilai}\: \: y=4,\\ &\qquad\textrm{maka sampai langkah di sini hanya }\\ &\qquad\textrm{ada 1 persamaan yang memenuhi}\\ &\qquad\textrm{yaitu}:\: \: y=\left | -2x+4 \right | \end{aligned} \end{array}$

$\begin{array}{l}\\ 15.&\textrm{Gambarlah garfik untuk persamaan}\: \: \left | x \right |+\left | y \right |=4\\\\\\ &\textrm{Jawab}:\\\\ &\textrm{untuk}\: \: \left | x \right |+\left | y \right |=4\\\\ &\begin{array}{|cc|cc|}\hline x> 0\: ,\: y> 0&&&x> 0\: ,\: y<0\\\hline x+y=4&&&x+(-y)=4\\ &&&\\\hline &&&\\ x<0\: ,\: y> 0&&&x<0\: ,\: y<0\\\hline (-x)+y=4&&&(-x)+(-y)=4\\\hline \end{array} \end{array}$

$\color{blue}\begin{aligned}&\qquad\textrm{berikut gambar grafiknya} \end{aligned}$

Contoh Soal Lanjutan Nilai Mutlak

$\begin{array}{ll}\\ 6.&\textrm{Perhatikanlah gambar berikut}\\ \end{array}$

$\begin{array}{ll}\\ &\textrm{Persamaan yang sesuai dengan grafik di atas adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\left | x-3 \right |\\ \textrm{b}.&y=-\left | x-3 \right |\\ \color{red}\textrm{c}.&y=\left | x+3 \right |\\ \textrm{d}.&y=\left | x+6 \right |\\ \textrm{e}.&y=-\left | x+6 \right |\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ \end{array}$

$\begin{array}{l}\\ 7.&\textrm{Diketahui suatu fungsi}\: \: y=\left | 2x \right |\\ &\textrm{Nilai saat}\: \: -2\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&-2\\ \textrm{c}.&0\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&4\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}y&=\left | 2x \right |\\ \textrm{saat}&\: \: x=-2,\: \: \textrm{maka}\\ y&=\left | 2(-2) \right |\\ &=\left | -4 \right |\\ &=-(-4)\\ &=4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 8.&\textrm{Diketahui suatu fungsi}\: \: f(x)=\left | x-2 \right |\\ &\textrm{Nilai}\: \: f(-2)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&4\\ \textrm{d}.&6\\ \textrm{e}.&8\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}f(x)&=\left | x-2 \right |\\ \textrm{Nilai}&\: \: f(-2)\: \: \textrm{berarti nilai}\\ \textrm{saat}&\: \: x=-2,\: \: \textrm{maka}\\ f(x)&=\left | x-2 \right |\\ f(-2)&=\left | -2-2 \right |\\ &=\left | -4 \right |\\ &=-(-4)\\ &=4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 9.&\textrm{Diketahui fungsi}\: \: f(x)=\left | 2-4x\right |\\ &\textrm{Nilai dari}\: \: f(1)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \textrm{b}.&0\\ \textrm{c}.&1\\ \color{red}\textrm{d}.&2\\ \textrm{e}.&4\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}f(x)&=\left | 2-4x \right |\\ \textrm{Nilai}&\: \: f(1)\: \: \textrm{berarti nilai}\\ \textrm{saat}&\: \: x=1,\: \: \textrm{maka}\\ f(x)&=\left | 2-4x) \right |\\ f(1)&=\left | 2-4(1) \right |\\ &=\left | -2 \right |\\ &=-(-2)\\ &=2 \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&\textrm{Nilai}\: \: k\: \: \textrm{yang memenuhi}\: \: \left | 4k \right |=16\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&\pm 2\\ \color{red}\textrm{d}.&\pm 4\\ \textrm{e}.&\pm 8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left | 4k \right |&=16\\ (4k)&=\pm 16\\ k&=\pm \displaystyle \frac{16}{4}\\ &=\pm 4 \end{aligned}\end{array}$

Contoh Soal Nilai Mutlak (Kelas X Matematika Wajib)

$\begin{array}{ll}\\ 1.&\textrm{Penyelesaian}\: \: -5\left |x-7 \right |+2=-13\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x=-10\: \: \textrm{dan}\: \: x=-4\\ \textrm{b}.&x=-10\: \: \textrm{dan}\: \: x=-2\\ \textrm{c}.&x=-2\: \: \textrm{dan}\: \: x=4\\ \textrm{d}.&x=4\: \: \textrm{dan}\: \: x=-10\\ \color{red}\textrm{e}.&x=4\: \: \textrm{dan}\: \: x=10\end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}-5\left |x-7 \right |&=-13-2\\ \left |x-7 \right |&=\displaystyle \frac{-15}{-5}\\ \left |x-7 \right |&=3\\ x-7&=\pm 3\\ x&=\mp 3+7\\ x&=\begin{cases} +3 & \text{ + } 7=10 \\\\ -3 & \text{ + } 7=4 \end{cases} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui persamaan}\: \: \left |x-5 \right |+\left | x+9 \right |=22\\ &\textrm{dan diberikan pernyataan-pernyataan berikut}\\ &\textrm{untuk}\\ &(\textrm{i})\quad x<-9,\: \textrm{maka}\: \: -x+5-x+9=22\\ &(\textrm{ii})\: \: \: -9\leq x< 5,\: \textrm{maka}\: \: -x+5-x+9=22\\ &(\textrm{iii})\: \: x<-9,\: \textrm{maka}\: \: -x+5-x-9=22\\ &(\textrm{iv})\: \: \: x\geq 5,\: \textrm{maka}\: \: x-5+x+9=22\\ &\textrm{Pernyataan yang tepa ditunjukkan oleh}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{ii})\\ \textrm{b}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \color{red}\textrm{c}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iv})\\ \textrm{d}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{e}.&(\textrm{iii})\: \: \textrm{dan}\: \: (\textrm{iv}) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned} \end{aligned}\end{array}$

$\begin{array}{l}\\ 3.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &3\left |x-1 \right |-2\left | x+1 \right |=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x=-\displaystyle \frac{1}{5}\: \: \textrm{dan}\: \: x=5\\ \textrm{b}.&x=-5\: \: \textrm{dan}\: \: x=-\displaystyle \frac{1}{5}\\ \color{red}\textrm{c}.&\displaystyle x=\frac{1}{5}\: \: \textrm{dan}\: \: x=5\\\\ \textrm{d}.&x=-5\: \: \textrm{dan}\: \: x=\displaystyle \frac{1}{5}\\\\ \textrm{e}.&x=-5\: \: \textrm{dan}\: \: x=5\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}3\left |x-1 \right |&-2\left | x+1 \right |=0 \\ 3\left |x-1 \right |&=2\left | x+1 \right |\\ 3^{2}(x-1)^{2}&=2^{2}(x+1)^{2}\\ 9\left ( x^{2}-2x+1 \right )&=4\left ( x^{2}+2x+1 \right )\\ 9x^{2}-4x^{2}&-18x-8x+9-4=0\\ 5x^{2}-26x&+5=0\\ (5x-1)&(x-5)=0\\ x=\displaystyle \frac{1}{5}\quad &\textrm{atau}\quad x=5 \end{aligned} \end{array}$

$\begin{array}{l}\\ 4.&\textrm{Himpunan penyelesaian dari}\: \: \left | 5-\displaystyle \frac{2}{3}x \right |-9=8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ -33,-18 \right \}\\ \color{red}\textrm{b}.&\left \{ -18,33 \right \}\\ \textrm{c}.&\left \{ 33,8 \right \}\\ \textrm{d}.&\left \{ 8,-33 \right \}\\ \textrm{e}.&\left \{ -8,33 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\left | 5-\displaystyle \frac{2}{3}x \right |&-9=8\\ \left | 5-\displaystyle \frac{2}{3}x \right |&=8+9=17\\ 5-\displaystyle \frac{2}{3}x&=\pm 17\\ -\displaystyle \frac{2}{3}x&=\pm 17-5\\ x&=\displaystyle \frac{\pm 17-5}{\left ( -\displaystyle \frac{2}{3} \right )}\\ x_{1}&=\displaystyle \left (\frac{17-5}{-2} \right )\times 3\\ &=-18\\ x_{2}&=\displaystyle \left (\frac{-17-5}{-2} \right )\times 3\\ &=33 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Harga buku di toko "KITA" adalah}\: \: 28.000,-\\ &\textrm{Jika harga buku tulis di toko "FAMILI" memiliki selisih}\: \: 7.000,-\\ &\textrm{Harga buku tulis di toko "FAMILI" dapat dinyatakan dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left | x-7000 \right |=28000\\ \textrm{b}.&\left | x+7000 \right |=28000\\ \color{red}\textrm{c}.&\left | x-28000 \right |=7000\\ \textrm{d}.&\left | x+28000 \right |=7000\\ \textrm{e}.&\left | x-21000 \right |=7000\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ \end{array}$

Contoh Soal Lanjutan 4 Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 21.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\tan 3x}{1-\cos ^{2}2x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -\frac{3}{2}\\\\ \textrm{b}.&\displaystyle -\frac{3}{4}\\\\ \textrm{c}.&\displaystyle -\frac{1}{2}\\\\ \textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array} \end{array}$

Jawab : e

Berikut pembahasannya

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\tan 3x}{1-\cos ^{2}2x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\tan 3x}{\sin ^{2}2x}\\ &=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\times \tan \textcircled{3}x}{\sin \textcircled{2}x\times \sin \textcircled{2}x}\\ &=\displaystyle \frac{1\times 3}{2\times 2}\\ &=\displaystyle \frac{3}{4} \end{aligned}$

$\begin{array}{ll}\\ 22.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-9}{\sin (x-3)}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle 9\\\\ \textrm{b}.&\displaystyle 7\\\\ \textrm{c}.&\displaystyle 6\\\\ \textrm{d}.&\displaystyle 3\\\\ \textrm{e}.&\displaystyle -6 \end{array} \end{array}$

Jawab : c

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-9}{\sin (x-3)}&\quad \textrm{perhatikan bahwa}\\ &\color{black}\textrm{misal}\: \: x=3+h,\: \: \textrm{maka}\: \: h\rightarrow 0\\ &\color{red}\textrm{selanjutnya gunakan}\: \: h=x-3,\: \: \textrm{dengan}\: \: h\rightarrow 0\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{(3+h)^{2}-9}{\sin h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{h^{2}+6h+9-9}{\sin h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \:\frac{h(h+6)}{\sin h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \:(h+6)\\ &=0+6=6 \end{aligned}$

atau boleh juga

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-9}{\sin (x-3)}&=\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{(x+3)(x-3)}{\sin (x-3)}\\ &=\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle (x+3)\\ &=3+3=6 \end{aligned}$

$\begin{array}{ll}\\ 23.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{x-\displaystyle \frac{\pi }{2}}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle 6\\\\ \textrm{b}.&\displaystyle 3\\\\ \textrm{c}.&\displaystyle 2\\\\ \textrm{d}.&\displaystyle -2\\\\ \textrm{e}.&\displaystyle -6 \end{array} \end{array}$

Jawab : e

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{x-\displaystyle \frac{\pi }{2}}&\quad \color{black}\textrm{perhatikan juga bahwa}\\ &x=\displaystyle \frac{\pi }{2}+h,\quad \textrm{dengan}\: \: \: h\rightarrow 0\\ &\color{red}\textrm{atau}\: \: \: h=x-\displaystyle \frac{\pi }{2}\quad \textrm{dengan}\: \: \: h\rightarrow 0\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6\left ( \displaystyle \frac{\pi }{2}+h \right )}{h}=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( \displaystyle 3\pi +6h \right )}{h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( \displaystyle \pi +6h \right )}{h}=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{-\sin \left ( \displaystyle 6h \right )}{h}\\ &=-6 \end{aligned}$

Daftar Pustaka

1. Astuti, A. N., Miyanto, dan Ngapiningsih. 2020. Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII. Yogyakarta: PT. PENERBIT INTAN PARIWARA.