Contoh Soal 4 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 16.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x-4}<\sqrt{2x+8}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -4<x\le {\displaystyle \frac{2}{3}}\\ \text{b}.& -4<x<3\\ \text{c}.& {\displaystyle \frac{2}{3}\le x<3}\\ \text{d}.& 2<x\le 4\\ \text{e}.& -4\le x\le 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \sqrt{6x-4}}& <\sqrt{2x+8}\\ 1.\text{Kuadrat}& \text{kan}\\ 6x-4& <2x+8\\ 6x-2x& <8+4\\ 4x& <12\\ x& <3\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x-4& \ge 0\\ 6x& \ge 4\\ x& \ge {\displaystyle \frac{2}{3}}\\ 3.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 2x+8& \ge 0\\ 2x& \ge -8\\ x& \ge -4\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{x+3}>\sqrt{12-2x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 3<x\le 6\\ \text{b}.& -3<x\le 6\\ \text{c}.& -6<x\le 3\\ \text{d}.& -6<x\le -3\\ \text{e}.& x<3\text{atau}x>6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle {\displaystyle \sqrt{x+3}}}& >\sqrt{12-2x}\\ 1.\text{Kuadrat}& \text{kan}\\ x+3& >12-2x\\ x+2x& >12-3\\ 3x& >9\\ x& >3\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ x+3& \ge 0\\ x& \ge -3\\ 3.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 12-2x& \ge 0\\ 2x-12& \le 0\\ 2x& \le 12\\ x& \le 6\end{array}\end{array}$

$\begin{array}{l}\\ 18.& (\mathbf{\text{SBMPTN 2013 Mat Das}})\\ & \text{Jika}1<m<2,\text{maka semua nilai}\\ & x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2+4x}{-x^2+3x-3m}>0\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x>-3\\ \text{b}.& x<-4\\ \text{c}.& -4<x<0\\ \text{d}.& x<-4\text{atau}x>0\\ \text{e}.& x<-3\text{atau}x>-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}1.& \text{Diketahui bahwa}:{\displaystyle \frac{x^2+4x}{-x^2+3x-3m}>0}\\ & \text{dengan kondisi}1<m<2\\ & \text{Perhatikanlah penyebutnya yang}\\ & \text{mengandung bilangan}m\text{yang terletak}\\ & \text{pada interval}:1<m<2.\\ 2.& \text{Kita cek kondisi penyebutnya dengan}\\ & \text{menentukan}Diskriminan(D)-\text{nya}\\ & \text{yaitu}:\\ & a{x}^2+bx+c\{\begin{array}{l}a>0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit positif}\\ a<0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit negatif}\end{array}\\ & \text{Karena penyebutnya}:-x^2+3x-3m,\\ & \text{dengan}a=-1,b=3,\mathrm{\&}c=-3m,\text{maka}\\ & D=3^2-4(-1)(-3m)=9-12m\\ 3.& \text{Karena nilai}m\text{berada pada}1<m<2\\ & \text{maka}\\ & 1<m<2\\ & \iff 12<12m<24\\ & \iff -12>-12m>-24\\ & \iff 9-12>9-12m>-13\\ & \iff -3>9-12m>-13\\ & \iff -13<9-12m<-3\\ & \text{Ini berarti nilai}D\text{negatif, sehingga}\\ & \text{berakibat penyebut berupa}-x^2+3x-3m\\ & \text{adalah wilayah}definitnegatif\\ 4.& \text{Selanjutnya pemfaktoran pertidaksamaan}\\ & \bullet \text{semula}\\ & {\displaystyle \frac{x(x+4)}{\underset{definitnegatif}{\underbrace{-x^2+3x-3m}}}>0\iff \frac{x(x+4)}{-}>0}\\ & \bullet \text{akan berubah menjadi}\\ & x(x+4)<0\\ & \text{pembuat nol-nya adalah}:x(x+4)=0\\ & \text{maka}x=-4\text{atau}x=0,\text{sehingga}\\ & \text{interval nilai}x-\text{nya}:-4<x<0\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Jika}1<a<2,\text{maka semua nilai}\\ & x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{-x^2+2ax-6}{x^2+3x}\le 0\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x<-3\text{atau}x>0\\ \text{b}.& x<-3\text{atau}x\ge -2\\ \text{c}.& x\le -2\text{atau}x\ge 2\\ \text{d}.& -3<x<0\\ \text{e}.& -2\le x<0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}1.& \text{Diketahui bahwa}:{\displaystyle \frac{-x^2+2ax-6}{x^2+3x}\le 0}\\ & \text{untuk membedakan}a\text{pada persamaan}\\ & \text{kuadrat dengan}a\text{di atas, selanjutnya}\\ & \text{kita menuliskan}a\text{di atas dengan}:m\\ & \text{karena}1<a<2\text{diubah}:1<m<2\\ & \text{Perhatikanlah pembilang yang}\\ & \text{mengandung bilangan}m\text{yang terletak}\\ & \text{pada interval}:1<m<2.\\ 2.& \text{Kita cek kondisi pembilangnya dengan}\\ & \text{menentukan}Diskriminan(D)-\text{nya}\\ & \text{yaitu}:\\ & a{x}^2+bx+c\{\begin{array}{l}a>0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit positif}\\ a<0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit negatif}\end{array}\\ & \text{Karena pebilangnya}:-x^2+2mx-6,\\ & \text{dengan}a=-1,b=2m,\mathrm{\&}c=-6,\text{maka}\\ & D=(2m{)}^2-4(-1)(-6)=4m^2-24\\ 3.& \text{Karena nilai}m\text{berada pada}1<m<2\\ & \text{maka}\\ & 1<m<2\\ & \iff 1^2<m^2<2^2\iff 1<m^2<4\\ & \iff 4<4m^2<16\\ & \iff 4-24<4m^2-24<16-24\\ & \iff -20<4m^2-24<-8\\ & \text{Ini berarti nilai}D\text{negatif, sehingga}\\ & \text{berakibat pembilangnya berupa}-x^2+2mx-6\\ & \text{adalah wilayah}definitnegatif\\ 4.& \text{Selanjutnya pemfaktoran pertidaksamaan}\\ & \bullet \text{semula}\\ & {\displaystyle \frac{\stackrel{definitnegatif}{\overbrace{-x^2+2mx-6}}}{x(x+3)}\le 0\iff \frac{-}{x(x+3)}\le 0}\\ & \bullet \text{akan berubah menjadi}\\ & x(x+3)>0\\ & \text{pembuat nol-nya adalah}:x(x+3)=0\\ & \text{maka}x=-3\text{atau}x=0,\text{sehingga}\\ & \text{interval nilai}x-\text{nya}:x<-3\text{atau}x>0\end{array}\end{array}$

$\begin{array}{l}\\ 20.& (\text{KSM 2019})\\ & \text{Banyaknya selesaian real dari persamaan}\\ & 2x^2-7x+6=15\lfloor {\displaystyle \frac{1}{x}}\rfloor \lfloor x\rfloor \\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{tidak ada}}\\ & \end{array}$

$.\begin{array}{rl}\text{Solu}& \text{si jawaban merujuk dari blog Pak Anang}\\ \text{And}& \text{a bisa mengunjunginya di blognya beliau}\end{array}$

$.\begin{array}{rl}\text{di}& \end{array}$  sini (Soal dan Pembahasan KSM 2019 oleh Pak Anang)

Contoh Soal 3 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 11.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x-5}\le x\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 1<x<0\\ \\ \text{b}.& x<1\text{atau}x\ge 5\\ \\ \text{c}.& {\displaystyle \frac{5}{6}\le x\le 1\text{atau}x\ge 5}\\ \\ \text{d}.& {\displaystyle \frac{5}{6}\le x<1\text{atau}5<x<6}\\ \\ \text{e}.& x\ge 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{6x-5}& \le x\\ 1.\text{Kuadrat}& \text{kan}\\ 6x-5& \le x^2\\ -x^2+6x-5& \le 0\\ x^2-6x+5& \ge 0\\ (x-1)(x-5)& \ge 0\\ x\le 1\text{atau}& x\ge 5\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x-5& \ge 0\\ 6x& \ge 5\\ x& \ge {\displaystyle \frac{5}{6}}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x+6}>6\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x>7\\ \text{b}.& x\ge 7\\ \text{c}.& x<7\\ \text{d}.& x>1\\ \text{e}.& x\ge 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\sqrt{6x+6}& >6\\ 1.\text{Kuadrat}& \text{kan}\\ 6x+6& >36\\ x+1& >6\\ x& >7\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x+6& \ge 0\\ 6x& \ge 6\\ x& \ge {\displaystyle \frac{6}{6}}\\ x& \ge 1\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Penyelesaian pertidaksamaan}\\ & x+2>{\displaystyle \sqrt{10-x^2}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\le x\le \sqrt{10}\\ \text{b}.& 1<x\le \sqrt{10}\\ \text{c}.& -3<x\le \sqrt{10}\\ \text{d}.& -\sqrt{10}\le x\le \sqrt{10}\\ \text{e}.& x<-3\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}x+2& >{\displaystyle \sqrt{10-x^2}}\\ 1.\text{Kuadrat}& \text{kan}\\ x^2+4x+4& >10-x^2\\ 2x^2+4x& +4-10>0\\ 2x^2+4x-& 6>0\\ x^2+2x-& 3>0\\ (x+3)& (x-1)>0\\ x<-3& \text{atau}x>1\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 10-x^2& \ge 0\\ x^2-10& \le 0\\ (x-\sqrt{10})& (x+\sqrt{10})\le 0\\ -\sqrt{10}\le x& \le \sqrt{10}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{3x+7}\ge x-1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -1<x<6\\ \text{b}.& -1\le x<6\\ \text{c}.& x\ge -{\displaystyle \frac{7}{3}}\\ \text{d}.& -{\displaystyle \frac{7}{3}\le x\le 6}\\ \text{e}.& -{\displaystyle \frac{7}{3}\le x\le 1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{3x+7}& \ge x-1\\ 1.\text{Kuadrat}& \text{kan}\\ 3x+7& \ge x^2-2x+1\\ -x^2+3x& +2x+7-1\ge 0\\ -x^2+5x+& 6\ge 0\\ x^2-5x-& 6\le 0\\ (x+1)& (x-6)\le 0\\ -1\le x& \le 6\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 3x+7& \ge 0\\ 3x& \ge -7\\ x& \ge -{\displaystyle \frac{7}{3}}\end{array}\end{array}$

 $\begin{array}{l}\\ 15.& \text{Nilai}x\text{yang memenuhi}\\ & \sqrt{2x-8}<\sqrt{x+5}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x\ge -5\\ \text{b}.& x<-13\text{atau}x\ge -4\\ \text{c}.& x<13\\ \text{d}.& 4\le x<13\\ \text{e}.& -5\le x\le 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{2x-8}& <\sqrt{x+5}\\ (1)\text{kuadratkan}& \\ 2x-8& <x+5\\ x& <13\\ (2)2x-8\ge 0& \\ x& \ge 4\\ (3)x+5\ge 0& \\ x& \ge 5\\ \text{perhatikan}& \text{lah garis bilangannya berikut}\\ 1& \begin{array}{c}\\ & & & & & & & & & & \\ \text{cline}1-7& & & & & & & & & & \text{X}\\ & & & & & & 13& & & & \end{array}\\ \\ 2& \begin{array}{c}\\ & & & & & & & & & & \\ \text{cline}4-11& & & & & & & & & & \text{X}\\ & & & 4& & & & & & & \end{array}\\ \\ 3& \begin{array}{c}\\ & & & & & & & & & & \\ \text{cline}6-11& & & & & & & & & & \text{X}\\ & & & & & 5& & & & & \end{array}\end{array}\end{array}$

Contoh Soal 2 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

 $\begin{array}{l}\\ 6.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{2x+6}{x-4}\le 1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -10<x<4\\ \text{b}.& -10\le x<4\\ \text{c}.& -4<x\le 10\\ \text{d}.& x\le -10\text{atau}x\ge 4\\ \text{e}.& x<-10\text{atau}x\ge 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{2x+6}{x-4}}& \le 1\\ {\displaystyle \frac{2x+6}{x-4}-1}& \le 0\\ {\displaystyle \frac{2x+6-(x-4)}{x-4}}& \le 0\\ {\displaystyle \frac{x+10}{x-4}}& \le 0\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Nilai}x\text{yang memenuhi}{\displaystyle \frac{x^2-x}{x+3}\ge 1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x<-3\text{atau}-1\le x\le 3\\ \text{b}.& -3<x\le -1\text{atau}x\ge 3\\ \text{c}.& -3\le x\le 3\\ \text{d}.& -3\le x\le -1\text{atau}x\ge 3\\ \text{e}.& -3\le x\le -1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2-x}{x+3}}& \ge 1\\ {\displaystyle \frac{x^2-x}{x+3}}& -1\ge 0\\ {\displaystyle \frac{x^2-x-(x+3)}{x+3}}& \ge 0\\ {\displaystyle \frac{x^2-2x-3}{x+3}}& \ge 0\\ {\displaystyle \frac{(x-3)(x+1)}{x+3}}& \ge 0\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Nilai}x\text{yang memenuhi}\\ & x+2+{\displaystyle \frac{1}{x+4}>0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x<-4\text{atau}x\ge -3\\ \text{b}.& x<-4\text{atau}x>-3\\ \text{c}.& -4\le x\le -3\\ \text{d}.& x>-4\\ \text{e}.& -4\le x\le -3\text{atau}x>-3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}x+2+{\displaystyle \frac{1}{x+4}}& >0\\ {\displaystyle \frac{(x+2)(x+4)+1}{(x+4)}}& >0\\ {\displaystyle \frac{x^2+6x+8+1}{x+4}}& >0\\ {\displaystyle \frac{x^2+6x+9}{x+4}}& >0\\ {\displaystyle \frac{(x+3{)}^2}{(x+4)}}& >0\\ x& >-4\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Nilai}x\text{yang memenuhi}\\ & x+3<{\displaystyle \frac{x^2+6x+11}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-3{\displaystyle \frac{2}{3}\text{atau}x>0,x\in \mathbb{R}}\}\\ \text{b}.& \{x|0\le x\le 11,x\in \mathbb{R}\}\\ \text{c}.& \{x|x<-11\text{atau}x>0,x\in \mathbb{R}\}\\ \text{d}.& \{x|x<0\text{atau}x>11,x\in \mathbb{R}\}\\ \text{e}.& \{x|x\le 11\text{atau}x>0,x\in \mathbb{R}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}x+3<{\displaystyle \frac{x^2+6x+11}{x}}& \\ x+3-{\displaystyle \frac{x^2+6x+11}{x}}& <0\\ {\displaystyle \frac{x(x+3)-(x^2+6x+11)}{x}}& <0\\ {\displaystyle \frac{x^2+3x-x^2-6x-11}{x}}& <0\\ {\displaystyle \frac{-3x-11}{x}}& <0\\ {\displaystyle \frac{3x+11}{x}}& >0\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Nilai}x\text{berikut yang tidak memenuhi}\\ & {\displaystyle \frac{x-3}{x^2+2x+1}\le 0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -1\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{x-3}{x^2+2x+1}\le 0}& \\ {\displaystyle \frac{(x-3)}{(x+1{)}^2}\le 0}& \\ \text{Pembuat nol}& \\ \{\begin{array}{ll}x& =3,\text{boleh digunakan}\\ x& =-1,\text{textrm{tetapi}}x\ne -1,\end{array}& \\ \text{sehingga}-1\text{tidak digunakan}& \\ \begin{array}{c}\\ & & & & & & & & & & \\ & -& -& -& -& -& & +& +& & \\ & & -1& & & & 3& & & & \\ & & & & & & & & & & \end{array}& \end{array}\end{array}$

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

$\begin{array}{l}\\ 1.& \text{Diketahui pertidaksamaan}{\displaystyle \frac{x+10}{x-9}\le 0}\\ & \text{dan diberikan beberapa nilai berikut}\\ & (\text{i})x=-6(\text{iii})x=-14\\ & (\text{ii})x=-10(\text{iv})x=-18\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \text{di atas adalah ditunjukkan oleh}....\\ & \begin{array}{l}\\ \text{a}.& (\text{i})\text{dan}(\text{ii})\\ \text{b}.& (\text{i})\text{dan}(\text{iii})\\ \text{c}.& (\text{ii})\text{dan}(\text{iii})\\ \text{d}.& (\text{ii})\text{dan}(\text{iv})\\ \text{e}.& (\text{iii})\text{dan}(\text{iv})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{x+10}{x-9}}& \le 0\\ \text{HP}=& \{x|-10\le x<9,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\text{ adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 2\le x<6\\ \text{b}.& 2\le x<3\\ \text{c}.& 2<x<3\text{atau}x>6\\ \text{d}.& x<3\text{atau}3<x<6\\ \text{e}.& x<2\text{atau}x>3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \frac{6}{x-3}<\frac{8}{x-2}}& \\ {\displaystyle \frac{6}{x-3}-\frac{8}{x-2}}& <0\\ {\displaystyle \frac{6(x-2)-8(x-3)}{(x-3)(x-2)}}& <0\\ {\displaystyle \frac{6x-8x-12+24}{(x-2)(x-3)}}& <0\\ {\displaystyle \frac{-2x+12}{(x-2)(x-3)}}& <0\\ {\displaystyle \frac{2x-12}{(x-2)(x-3)}}& >0\\ {\displaystyle \frac{2(x-6)}{(x-2)(x-3)}}& >0\\ \text{HP}=& \{x|2<x<3\text{atau}x>6,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \frac{x^2-81}{x^2}\ge 0\text{ adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x\le -9\text{atau}x\ge 9\\ \text{b}.& -9\le x<0\text{atau}x\ge 9\\ \text{c}.& -9\le x<0\text{atau}0<x\le 9\\ \text{d}.& -9<x\le 9\\ \text{e}.& x\in \mathbb{R}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2-81}{x^2}}& \ge 0\\ {\displaystyle \frac{(x+9)(x-9)}{x^2}}& \ge 0\\ \text{HP}=& \{x|x\le -9\text{atau}x\ge 9,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2-4}{x+2}>0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x>2\\ \text{b}.& -2\le x<2\\ \text{c}.& x<-2\text{atau}x>2\\ \text{d}.& x<-2\text{atau}-2<x<2\\ \text{e}.& x\ge -2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2-4}{x+2}}& >0\\ {\displaystyle \frac{(x+2)(x-2)}{(x+2)}}& >0\\ (x-2)& >0\\ x& >2\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2+x-30}{2x^2+13x-45}<0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& \{x|-9<x<5,x\in \mathbb{R}\}\\ \text{b}.& \{x|-6<x<5,x\in \mathbb{R}\}\\ \text{c}.& \{x|-9<x<-6\text{atau}x<5,x\in \mathbb{R}\}\\ \text{d}.& \{x|-9<x<-6\text{atau}{\displaystyle \frac{5}{2}<x<5,x\in \mathbb{R}}\}\\ \text{e}.& \{x|x<-9\text{atau}-6<x<{\displaystyle \frac{5}{2}\text{atau}x<5,x\in \mathbb{R}}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2+x-30}{2x^2+13x-45}}& <0\\ {\displaystyle \frac{(x+6)(x-5)}{(x+9)(2x-5)}}& <0\\ \text{Cukup jelas}& \end{array}\end{array}$

Pertidaksamaan Rasional dan Irasional Satu Variabel

 A. Bentuk Umum Pertidaksamaan Rasional

$\begin{array}{rl}& \mathbf{\text{Pertidak samaan Rasional}}\\ & \\ & \{\begin{array}{ll}\mathbf{\text{A}}& \begin{array}{rl}& \\ \{\begin{array}{ll}{\displaystyle \frac{f(x)}{g(x)}}& <0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \le 0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& >0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \ge 0\end{array}& \begin{array}{cc}\mathbf{\text{misalnya}}& \{\begin{array}{ll}{\displaystyle \frac{x-1}{x+3}}& <0\\ \\ {\displaystyle \frac{x-1}{x+3}}& \le 0\\ \\ {\displaystyle \frac{x-1}{x+3}}& >0\\ \\ {\displaystyle \frac{x-1}{x+3}}& \ge 0\end{array}\end{array}\\ & \end{array}\\ \mathbf{\text{B}}& \begin{array}{rl}& \\ \{\begin{array}{ll}{\displaystyle \frac{f(x)}{g(x)}}& <0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \le 0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& >0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \ge 0\end{array}& \begin{array}{cc}\mathbf{\text{misalnya}}& \{\begin{array}{ll}{\displaystyle \frac{x^2-4}{x+6}}& <0\\ \\ {\displaystyle \frac{x^2-4}{x+6}}& \le 0\\ \\ {\displaystyle \frac{x^2-4}{x+6}}& >0\\ \\ {\displaystyle \frac{x^2-4}{x+6}}& \ge 0\end{array}\end{array}\\ & \end{array}\\ \mathbf{\text{C}}& \begin{array}{rl}& \\ \{\begin{array}{ll}{\displaystyle \frac{f(x)}{g(x)}}& <0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \le 0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& >0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \ge 0\end{array}& \begin{array}{cc}\mathbf{\text{misalnya}}& \{\begin{array}{ll}{\displaystyle \frac{x^2+2x-3}{x^2-4}}& <0\\ \\ {\displaystyle \frac{x^2+2x-3}{x^2-4}}& \le 0\\ \\ {\displaystyle \frac{x^2+2x-3}{x^2-4}}& >0\\ \\ {\displaystyle \frac{x^2+2x-3}{x^2-4}}& \ge 0\end{array}\end{array}\\ & \end{array}\end{array}\\ & \end{array}$

B. Menyelesaikan Pertidaksamaan Rasional

Misal pada bentuk A di atas, maka penyelesaiannya adalah

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

$$\begin{gathered} \mathbf{\text{Bentuk Umum}}1: \\ \sqrt{f(x)}\cdots A\{\begin{array}{l}\sqrt{f(x)}<A\\ \\ \sqrt{f(x)}\le A\\ \\ \sqrt{f(x)}>A\\ \\ \sqrt{f(x)}\ge A\end{array} \end{gathered}$$

Bentuk kedua

$$\begin{gathered} \mathbf{\text{Bentuk Umum}}2: \\ \sqrt{f(x)}\cdots \sqrt{g(x)}\{\begin{array}{l}\sqrt{f(x)}<\sqrt{g(x)}\\ \\ \sqrt{f(x)}\le \sqrt{g(x)}\\ \\ \sqrt{f(x)}>\sqrt{g(x)}\\ \\ \sqrt{f(x)}\ge \sqrt{g(x)}\end{array} \end{gathered}$$

D. Penyelesaian pada pertidaksamaan irasional

1. Kuadartakan masing-masing ruas
2. Dibawah tanda akar (numerus) haruslah $\ge 0$
3. Himpunan penyelesaian berupa irisan dari penyelesaian yang didapatkan

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

$\begin{array}{l}\\ 11.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{5}{4x-3}}\right|\le 1\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\le x\le \frac{3}{4}\text{atau}x\ge 2}\\ \text{b}.& x\le -{\displaystyle \frac{1}{2}\text{atau}{\displaystyle \frac{3}{4}<x\le 2}}\\ \text{c}.& -{\displaystyle \frac{1}{2}\le x\le 2,x\ne \frac{3}{4}}\\ \text{d}.& x\le -{\displaystyle \frac{1}{2}\text{atau}x>\frac{3}{4}}\\ \text{e}.& x\le -{\displaystyle \frac{1}{2}\text{atau}x\ge 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{5}{4x-3}}\right|& \le 1\\ -1\le {\displaystyle \frac{5}{4x-3}}& \le 1,\mathbf{\text{jika dibalik}}\\ -1\ge {\displaystyle \frac{4x-3}{5}}& \ge 1,\mathbf{\text{bentuk ini tidak}}\\ \mathbf{\text{dibolehkan}}& \mathbf{\text{maka perlu diubah menjadi}}\\ -1\ge {\displaystyle \frac{4x-3}{5}\text{atau}}& {\displaystyle \frac{4x-3}{5}\ge 1,\text{selanjutnya}}\\ \bullet \text{bagian}& 1\\ -1& \ge {\displaystyle \frac{4x-3}{5}\iff \frac{4x-3}{5}\le -1}\\ 4x-3& \le -5\\ 4x& \le -2\\ x& \le -{\displaystyle \frac{1}{2}}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{4x-3}{5}}& \ge 1\\ 4x-3& \ge 5\\ 4x& \ge 8\\ x& \ge 2\end{array}\end{array}$

$\begin{array}{l}\\ 12.& (\text{UMPTN 95})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{2}{2x-1}}\right|>1\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x>2\\ \text{b}.& x<2\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{c}.& x<-1\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{d}.& -1<x<2\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{e}.& x<-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{semua opsi bukan jawaban}}\\ & \mathbf{\text{Berikut pembahasannya}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{2}{2x-1}}\right|& >1\\ -1>{\displaystyle \frac{2}{2x-1}}& \text{atau}{\displaystyle \frac{2}{2x-1}>1,\mathbf{\text{dibalik}}}\\ -1<{\displaystyle \frac{2x-1}{2}}& \text{atau}{\displaystyle \frac{2x-1}{2}<1}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{2x-1}{2}}& >-1\\ 2x-1& >-2\\ 2x& >-1\\ x& >-{\displaystyle \frac{1}{2}}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{2x-1}{2}}& <1\\ 2x-1& <2\\ 2x& <3\\ x& <{\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& (\text{UMPTN 00})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{2x+7}{x-1}}\right|\ge 1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2\le x\le 8\\ \text{b}.& x\le -8\text{atau}x\ge -2\\ \text{c}.& -8\le x<1\text{atau}x>1\\ \text{d}.& -2\le x<1\text{atau}1<x\le 8\\ \text{e}.& x\le -8\text{atau}-2\le x<1\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{2x+7}{x-1}}\right|& \ge 1\\ -1\ge {\displaystyle \frac{2x+7}{x-1}}& \text{atau}{\displaystyle \frac{2x+7}{x-1}\ge 1}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{2x+7}{x-1}}& \le -1(\mathbf{\text{tidak boleh kali silang}})\\ {\displaystyle \frac{2x+7}{x-1}}& +1\le 0\\ & {\displaystyle \frac{2x+7+(x-1)}{x-1}\le 0}\\ {\displaystyle \frac{3x+6}{x-1}}& \le 0\\ {\text{HP}}_1=& \{x|-2\le x<1,x\in \mathbb{R}\}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{2x+7}{x-1}}& \ge 1\\ {\displaystyle \frac{2x+7}{x-1}}& -1\ge 0\\ & {\displaystyle \frac{2x+7-(x-1)}{x-1}\ge 0}\\ {\displaystyle \frac{x+8}{x-1}}& \ge 0\\ {\text{HP}}_2=& \{x|x\le -8\text{atau}x>1,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x\le -8\text{atau}-2\le x<1\text{atau}x>1,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{x-2}{x+3}}\right|\le 2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -8\le x<-3\\ \text{b}.& -8\le x<-1\\ \text{c}.& -4\le x<-3\\ \text{d}.& x\le -8\text{atau}x\ge -{\displaystyle \frac{4}{3}}\\ \text{e}.& x\le -4\text{atau}x\ge 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{x-2}{x+3}}\right|& \le 2\\ -2\le {\displaystyle \frac{x-2}{x+3}}& \text{atau}{\displaystyle \frac{x+2}{x+3}\le 2}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{x-2}{x+3}}& \ge -2(\mathbf{\text{tidak boleh kali silang}})\\ {\displaystyle \frac{x-2}{x+3}}& +2\ge 0\\ & {\displaystyle \frac{x-2+2(x+3)}{x+3}\ge 0}\\ {\displaystyle \frac{3x+4}{x+3}}& \ge 0\\ {\text{HP}}_1=& \{x|x<-3\text{atau}x\ge -{\displaystyle \frac{4}{3},x\in \mathbb{R}}\}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{x-2}{x+3}}& \le 2\\ {\displaystyle \frac{x-2}{x+3}}& -2\le 0\\ & {\displaystyle \frac{x-2-2(x+3)}{x+3}\le 0}\\ {\displaystyle \frac{-x-8}{x+3}}& \le 0,\mathbf{\text{koefisien textit{x} negatif}}\\ {\displaystyle \frac{x+8}{x+3}}& \ge 0\\ {\text{HP}}_2=& \{x|x\le -8\text{atau}x>-3,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x\le -8\text{atau}x\ge -{\displaystyle \frac{4}{3},x\in \mathbb{R}}\}\end{array}\end{array}$

$\begin{array}{l}\\ 15.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{2}{x+1}\le \left|x\right|\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& \{x|x\le -2\text{atau}x\ge 1\}\\ \text{b}.& \{x|x\le -2\text{atau}0\le x\le 1\}\\ \text{c}.& \{x|x\ge 1\}\\ \text{d}.& \{x|x<-1\text{atau}x\ge 1\}\\ \text{e}.& \{x|-1<x\le 1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left|x\right|& \ge {\displaystyle \frac{2}{x+1}\text{berakibat}}\\ {\displaystyle \frac{-2}{x+1}\ge x}& \text{atau}x\ge {\displaystyle \frac{2}{x+1}}\\ \bullet \text{bagian}& 1\\ x\le {\displaystyle \frac{-2}{x+1}}& (\mathbf{\text{tidak boleh kali silang}})\\ x+{\displaystyle \frac{2}{x+1}}& \le 0\\ & {\displaystyle \frac{x(x+1)+2}{x+1}\le 0}\\ {\displaystyle \frac{x^2+x+2}{x+1}}& \le 0\iff {\displaystyle \frac{\text{Definit positif}}{x+1}\le 0}\\ {\text{HP}}_1=& \{x|x<-1,x\in \mathbb{R}\}\\ \bullet \text{bagian}& 2\\ x& \ge {\displaystyle \frac{2}{x+3}}\\ x-& {\displaystyle \frac{2}{x+1}\ge 0}\\ & {\displaystyle \frac{x(x+1)-2}{x+1}\ge 0}\\ & {\displaystyle \frac{x^2+x-2}{x+1}\ge 0}\\ & {\displaystyle \frac{(x+2)(x-1)}{x+1}\ge 0}\\ {\text{HP}}_2=& \{x|-2\le x<-1\text{atau}x\ge 1,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x<-1\text{atau}x\ge 1,x\in \mathbb{R}\}\end{array}\end{array}$

Lanjutan Materi Pertidaksamaan Berkaitan Materi Pertidaksamaan Kuadrat

 Sebelumya 

silahkan buka di sini

A. PERTIDAKSAMAAN KUADRAT

$\begin{array}{l}\\ & \underset{―}{\mathbf{\text{Bentuk Umum}}}:\\ \\ & \{\begin{array}{l}a{x}^2+bx+c<0\\ a{x}^2+bx+c\le 0\\ a{x}^2+bx+c>0\\ a{x}^2+bx+c\ge 0\end{array}\\ & \end{array}$

B. PENYELESAIAN PERTIDAKSAMAAN KUADRAT

$\begin{array}{rl}& \\ a{x}^2+bx+c...0& \\ \text{diubah menjadi}& a{x}^2+bx+c=0\\ & \iff a(x-x_1)(x-x_2)=0\\ & \iff x=x_1\text{atau}x=x_2\\ & \end{array}$

$\begin{array}{|lc|}\hline \text{Pertidaksamaan}& \text{Himpunan Penyelesaian dengan}{x}_1<x_2\\ a{x}^2+bx+c<0& \{x|x_1<x<x_2,x\in \mathbb{R}\}\\ a{x}^2+bx+c\le 0& \{x|x_1\le x\le x_2,x\in \mathbb{R}\}\\ a{x}^2+bx+c>0& \{x|x<x_1\text{atau}x>x_2,x\in \mathbb{R}\}\\ a{x}^2+bx+c\ge 0& \{x|x\le x_1\text{atau}x\ge x_2,x\in \mathbb{R}\}\\ \hline\end{array}$

$\boxed{\text{CONTOH SOAL}}$

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

$\begin{array}{rl}& \\ x^2-6x+8...0& \\ \text{diubah menjadi}& x^2-6x+8=0\\ & \iff 1(x-2)(x-4)=0\\ & \iff x=2\text{atau}x=4\\ & \end{array}$

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

$\begin{array}{l}\\ 6.& \text{Penyelesaian dari pertidaksamaan}\\ & \left|{\displaystyle \frac{3-2x}{-5}}\right|>5\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-11\text{atau}x>14\\ \text{b}.& x<-14\text{atau}x>11\\ \text{c}.& 11<x<14\\ \text{d}.& -14<x<-11\\ \text{e}.& x>14\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{3-2x}{-5}}\right|& >5\\ {\displaystyle \frac{3-2x}{-5}<}& -5\text{atau}{\displaystyle \frac{3-2x}{-5}>5}\\ {\displaystyle \frac{2x-3}{5}>}& 5\text{atau}{\displaystyle \frac{2x-3}{5}<-5}\\ 2x-3>& 25\text{atau}2x-3<-25\\ 2x>25& +3\text{atau}2x<-25+3\\ x>14& \text{atau}x<-11,\\ & \text{dapat juga dituliskan}\\ x<-& 11\text{atau}x>14\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |2-2|x+1||>4\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-4\text{atau}x>2\\ \text{b}.& x<-3\text{atau}x>1\\ \text{c}.& x<-2\text{atau}x>0\\ \text{d}.& x<-1\text{atau}x>3\\ \text{e}.& x<0\text{atau}x>4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}|2-2|x+1||>4& \\ 2-2|x+1|<-4& \text{atau}2-2|x+1|>4\\ -2|x+1|<-6& \text{atau}-2|x+1|>2\\ |x+1|>3& \text{atau}|x+1|<-1\\ \{\begin{array}{c}(x+1)<-3\\ (x+1)>3\end{array}& \text{atau}\{\begin{array}{c}|x+1|<-1\\ \mathbf{\text{tak mungkin}}\end{array}\\ \text{Selanjutnya}& \text{akan didapatkan}\\ x<-4& \text{atau}x>2\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |3-\left|x\right||<10\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-14\text{atau}x>12\\ \text{b}.& x<-13\text{atau}x>13\\ \text{c}.& x<-12\text{atau}x>10\\ \text{d}.& 0<x<10\\ \text{e}.& -13<0<13\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}|3-\left|x\right||& <10\\ -10<3-& \left|x\right|<10\\ -13<-& \left|x\right|<7\\ -7<& \left|x\right|\le 13,(\text{ingat harga}\left|x\right|\ge 0)\\ 0\le & \left|x\right|<13\\ & \text{selanjutnya},\\ & \left|x\right|<13\\ -13<& x<13\\ \text{HP}=& \{x|-13<x<13,x\in \mathbb{R}\}\end{array}\end{array}$

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

$\begin{array}{l}\\ 10.& (\text{SPMB 05})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {|x-2|}^2<4|x-2|+12\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \{x\in \mathbb{R}|2\le x\le 8\}\\ \text{b}.& \{x\in \mathbb{R}|4<x<8\}\\ \text{c}.& \{x\in \mathbb{R}|-4<x<8\}\\ \text{d}.& \{x\in \mathbb{R}|-2<x<4\}\\ \text{e}.& \{x\in \mathbb{R}|2<x<4\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{misalkan}p& =|x-2|,\text{selanjutnya}\\ {|x-2|}^2<& 4|x-2|+12\\ p^2<& 4p+12\\ p^2-& 4p-12<0\\ (p-6)& (p+2)<0\\ -2<p& <6,\text{atau jika dikembalikan}\\ -2<& |x-2|<6,\text{ingat, nilanya tidak negatif}\\ 0\le & |x-2|<6\\ -6<& x-2<6\\ -4<& x<8\\ \text{HP}=& \{-4<x<8,x\in \mathbb{R}\}\end{array}\end{array}$

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

$\begin{array}{l}\\ 1.& \text{Nilai}x\text{yang memenuhi}\left|{\displaystyle \frac{1}{2}x+6}\right|\ge 9\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -12<x<6\\ \text{b}.& -30\le x\le 6\\ \text{c}.& x\ge 6\text{atau}x\le -30\\ \text{d}.& x<6\text{atau}x<-30\\ \text{e}.& x\le 6\text{atau}x\ge -30\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{1}{2}x+6}\right|& \ge 9\\ {\displaystyle \frac{1}{2}x+6}& \le -9\text{atau}{\displaystyle \frac{1}{2}x+6\ge 9}\\ {\displaystyle \frac{1}{2}x\le }& -9-6\text{atau}{\displaystyle \frac{1}{2}x\ge 9-6}\\ {\displaystyle \frac{1}{2}x\le }& -15\text{atau}{\displaystyle \frac{1}{2}x\ge 3}\\ x\le & -30\text{atau}x\ge 6\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & 3|x+1|\le |x-2|\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{4}\le x\le -\frac{1}{4}}\\ \text{b}.& -{\displaystyle \frac{5}{2}\le x\le \frac{5}{2}}\\ \text{c}.& x\le {\displaystyle \frac{1}{4}\text{atau}x\ge \frac{5}{2}}\\ \text{d}.& x\le -{\displaystyle \frac{5}{2}\text{atau}x\ge \frac{1}{4}}\\ \text{e}.& x\le -{\displaystyle \frac{5}{2}\text{atau}x\ge -\frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}3|x+1|& \le |x-2|\\ {\left(3|x+1|\right)}^2& \le {\left(|x-2|\right)}^2\\ {(3x+3)}^2& \le {(x-2)}^2\\ (3x+3+(x-2))& (3x+3-(x-2))\le 0\\ (4x+1)(2x+5)& \le 0\\ \text{HP}=& \{x|-{\displaystyle \frac{5}{2}\le x\le -\frac{1}{4},x\in \mathbb{R}}\}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |x-3|<3\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<3\\ \text{b}.& -3<x<3\\ \text{c}.& x<-3\text{atau}x<3\\ \text{d}.& x>0\text{atau}x<6\\ \text{e}.& x<0\text{atau}x<6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}|x-3|& <3\\ -3<(x-3)& <3\\ -3+3<x& <3+3\\ 0<x& <6\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |x+4|>8\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x>-8\\ \text{b}.& x<4\text{atau}x>12\\ \text{c}.& x>4\text{atau}x>-12\\ \text{d}.& x<-4\text{atau}x<6\\ \text{e}.& x>4\text{atau}x<-12\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}|x+4|& >8\\ (x+4)<-8& \text{atau}(x+4)>8\\ x<-12& \text{atau}x>4\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Himpunan penyelesaian dari pertidaksamaan}\\ & \left|{\displaystyle \frac{x+1}{2}}\right|>\left|{\displaystyle \frac{x-2}{3}}\right|\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \text{HP}=\{x|-7<x<{\displaystyle \frac{1}{5},x\in \mathbb{R}}\}\\ \text{b}.& \text{HP}=\{x|x<-7\text{atau}x>{\displaystyle \frac{1}{5},x\in \mathbb{R}}\}\\ \text{c}.& \text{HP}=\{x|x>-7,x\in \mathbb{R}\}\\ \text{d}.& \text{HP}=\{x|-1<x<2,x\in \mathbb{R}\}\\ \text{e}.& \text{HP}=\{x|x<-1\text{atau}x>2,x\in \mathbb{R}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{x+1}{2}}\right|& >\left|{\displaystyle \frac{x-2}{3}}\right|\\ {\left({\displaystyle \frac{x+1}{2}}\right)}^2& >{\left({\displaystyle \frac{x-2}{3}}\right)}^2\\ \left({\displaystyle \frac{x+1}{2}+\frac{x-2}{3}}\right)& \left({\displaystyle \frac{x+1}{2}-{\displaystyle \frac{x-2}{3}}}\right)>0\\ \left({\displaystyle \frac{3(x+1)+2(x-2)}{6}}\right)& \left({\displaystyle \frac{3(x+1)-2(x-2)}{6}}\right)>0\\ \left({\displaystyle \frac{5x-1}{6}}\right)& \left({\displaystyle \frac{x+7}{6}}\right)>0\\ \text{HP}=& \{x|x<-7\text{atau}x>{\displaystyle \frac{1}{5},x\in \mathbb{R}}\}\end{array}\end{array}$

Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

$\text{A. Sifat-Sifat yang Berlaku}$ 

Untuk a > 0, maka

berlaku sifat-sifat sebagai berikut:

$\{\begin{array}{ll}\left|x\right|<a& \iff -a<x<a\\ \left|x\right|\le a& \iff -a\le x\le a\\ \left|x\right|>a& \iff x<-a\text{atau}a>a\\ \left|x\right|\ge a& \iff x\le -a\text{atau}a\ge a\\ \\ \text{dan}& \text{perlu diingat}\\ \left|x\right|& =\sqrt{x^2}\end{array}$

$\text{B. Penyelesaian Pertidaksamaan}$

$\text{a. bentuk pertama}$

$|f(x)|\{\begin{array}{ll}<a& \Rightarrow -a\le f(x)\le a\\ \le a& \Rightarrow -a\le f(x)\le a\\ >a& \Rightarrow f(x)<-a\text{atau}f(x)>a\\ \ge a& \Rightarrow f(x)\le -a\text{atau}f(x)\ge a\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ \text{Tentukanlah himpunan penyelesaian dari}\\ (1).|3x-7|\le 11\\ (2).|3x-7|>11\\ \\ \text{Jawab}:\end{array}$

$\begin{array}{rl}(1).|3x-7|& \le 11\\ -11\le 3x-7& \le 11\\ -11+(7)\le 3x-7+(7)& \le 11+(7)\\ -4\le 3x& \le 18(\text{dibagi})3\\ -{\displaystyle \frac{4}{3}\le x}& \le 6\\ \text{HP}=& \{x|-{\displaystyle \frac{4}{3}\le x\le 6,x\in \mathbb{R}}\}\end{array}$

$\begin{array}{rl}(2).|3x-7|& >11\\ (3x-7)<-11& \text{atau}(3x-7)>11\\ 3x<-11+7& \text{atau}3x>11+7\\ x<-{\displaystyle \frac{4}{3}\text{atau}}& x>6\\ \text{HP}=& \{x|x<-{\displaystyle \frac{4}{3}\text{atau}x>6,x\in \mathbb{R}}\}\end{array}$

$\text{b. bentuk kedua}$

$|f(x)|\{\begin{array}{ll}<|g(x)|& \\ \le |g(x)|& \\ >|g(x)|& \\ \ge |g(x)|& \end{array}\Rightarrow \text{dikuadratkan kedua ruas}$

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

$\begin{array}{rl}a{x}^2+bx+c& \{\begin{array}{l}<\\ \le \\ >\\ \ge \end{array}0\\ \text{dengan}& a,b,c\in \mathbb{R},a\ne 0\end{array}$

dan juga

$\text{Akar-akar}\{\begin{array}{ll}{x}_1& \text{ dan }{x}_2\text{dengan}{x}_1\ne x_2\\ & \text{ serta}\\ x_1& \le x_2\end{array}$

maka dalam menentukan himpunan penyelesaiannya adalah sebagai berikut:

$\{\begin{array}{ll}a{x}^2+bx+c<0,& \text{HP}=\{x|x_1<x<x_2,x\in \mathbb{R}\}\\ a{x}^2+bx+c\le 0,& \text{HP}=\{x|x_1\le x\le x_2,x\in \mathbb{R}\}\\ a{x}^2+bx+c>0,& \text{HP}=\{x|x<x_1\text{atau}x>x_2,x\in \mathbb{R}\}\\ a{x}^2+bx+c\ge 0,& \text{HP}=\{x|x\le x_1\text{atau}x\ge x_2,x\in \mathbb{R}\}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ \text{Tentukanlah himpunan penyelesaian dari}\\ (1).|3x-7|\le |x-1|\\ (2).|3x-7|>|x-1|\\ \\ \text{Jawab}:\end{array}$

$\begin{array}{rl}(1).|3x-7|\le |x-1|& \\ {|3x-7|}^2\le {|x-1|}^2& \\ (3x-7{)}^2-(x-1{)}^2& \le 0\\ (3x-7+x-1)& (3x-7-(x-1))\le 0\\ (4x-8)(2x-6)& \le 0\\ (x-2)(x-3)& \le 0\\ \text{HP}=& \{x|2\le x\le 3,x\in \mathbb{R}\}\end{array}$

$\begin{array}{rl}(2).|3x-7|& >|x-1|\\ {|3x-7|}^2& >{|x-1|}^2\\ (3x-7{)}^2-& (x-1{)}^2>0\\ (3x-7+& x-1)(3x-7-(x-1))>0\\ (4x-8)& (2x-6)>0\\ (x-2)& (x-3)>0\\ \text{HP}=& \{x|x<2\text{atau}x>3,x\in \mathbb{R}\}\end{array}$

$\text{c. bentuk ketiga}$

Penyelesaian jenis ini mirip poin yang pertama, yaitu:

$\begin{array}{l}\\ |f(x)|\{\begin{array}{ll}\{\begin{array}{c}<g(x)\\ \le g(x)\end{array}& \text{ maka }\{\begin{array}{c}-g(x)<f(x)<g(x)\\ -g(x)\le f(x)\le g(x)\end{array}\\ \\ \{\begin{array}{c}>g(x)\\ \ge g(x)\end{array}& \text{ maka }\{\begin{array}{c}\{\begin{array}{c}f(x)>g(x)\\ f(x)<-g(x)\end{array}\\ \\ \{\begin{array}{c}f(x)\ge g(x)\\ f(x)\le -g(x)\end{array}\end{array}\end{array}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ \text{Tentukanlah himpunan penyelesaian dari}\\ (1).|3x-7|\le (x-1)\\ (2).|3x-7|>(x-1)\\ \\ \text{Jawab}:\end{array}$

$\begin{array}{rl}(1).-g(x)\le f(x)& \le g(x)\\ (\text{a}).\text{untuk}f(x)& \ge -g(x)\\ 3x-7& \ge -(x-1)\\ 4x& \ge 8\\ x& \ge 2\\ (\text{b}).\text{untuk}f(x)& \le g(x)\\ 3x-7& \le x-1\\ 2x& \le 6\\ x& \le 3\\ \text{HP}=& \{x|2\le x\le 3,x\in \mathbb{R}\}\end{array}$

$\begin{array}{rl}(2).f(x)<-g(x)& \text{atau}f(x)>g(x)\\ (\text{a}).\text{untuk}f(x)& <-g(x)\\ 3x-7& <-(x-1)\\ 4x& <8\\ x& <2\\ (\text{b}).\text{untuk}f(x)& >g(x)\\ 3x-7& >x-1\\ 2x& >6\\ x& >3\\ \text{HP}=& \{x|x<2\text{atau}x>3,x\in \mathbb{R}\}\end{array}$

Daftar Pustaka

1. Kanginan, M. 2016. Matematika untuk Siswa SMA-MA/SMK-MAK Kelas X (Wajib). Bandung: Srikandi Empat Widya Utama.
2. 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

$\begin{array}{l}\\ 21.& \text{Jumlah akar-akar persamaan}\\ & x^2+\left|x\right|-6=0,\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{x}^2+\left|x\right|& -6=0\\ {\left|x\right|}^2+\left|x\right|& -6=0\\ (\left|x\right|+3)& (\left|x\right|-2)=0\\ \left|x\right|=-3& \text{atau}\left|x\right|=2\\ \text{tidak meme}& \text{nuhi (-3) atau}x=\pm 2\\ & \{\begin{array}{ll}{x}_1& =2\\ x_2& =-2\end{array}\\ x_1+x_2& =2+(-2)=0\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Penyelesaian untuk}-2|x-5|=8\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \left\{0\right\}\\ \text{b}.& \left\{\right\}\\ \text{c}.& \{4,5\}\\ \text{d}.& \{1,9\}\\ \text{e}.& \{-1,9\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}-2|x-5|& =8\\ |x-5|& =-4\\ \text{Karena pe}& \text{rsamaan bernilai negatif},\\ \text{maka tida}& \text{k ada nilai}x\text{yang memenuhi}\\ \therefore \mathbf{\text{HP}}& =\left\{\right\}\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Nilai}x\text{yang memenuhi}\left|{\displaystyle \frac{x-5}{2x+1}}\right|=2\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{5}{3}\text{atau}\frac{3}{5}}\\ \text{b}.& {\displaystyle \frac{5}{3}\text{atau}-\frac{3}{5}}\\ \text{c}.& -{\displaystyle \frac{7}{3}\text{atau}\frac{3}{5}}\\ \text{d}.& {\displaystyle \frac{7}{3}\text{atau}-\frac{3}{5}}\\ \text{e}.& -{\displaystyle \frac{7}{3}\text{atau}\frac{3}{7}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{x-5}{2x+1}}\right|& =2\\ {\displaystyle \frac{x-5}{2x+1}}& =\pm 2\\ \text{untuk}& x=2\\ {\displaystyle \frac{x-5}{2x+1}}& =2\\ x-5& =2(2x+1)\\ x-4x& =2+5\\ -3x& =7\\ x& ={\displaystyle \frac{7}{-3}=-\frac{7}{3}}\\ \text{untuk}& x=-2\\ x-5& =-2(2x+1)\\ x+4x& =-2+5\\ 5x& =3\\ x& ={\displaystyle \frac{3}{5}}\end{array}\end{array}$

$\begin{array}{l}\\ 24.& \text{Nilai}x\text{yang memenuhi}||5x-4|-3|=2\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -3\text{atau}-6\\ \text{b}.& -3\text{atau}6\\ \text{c}.& 3\text{atau}-6\\ \text{d}.& 3\text{atau}6\\ \text{e}.& 6\text{atau}9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}||2x-3|-4|& =5\\ |2x-3|-4& =\pm 5\\ |2x-3|& =\pm 5+4\text{maka},\\ |2x-3|& =5+4=9(\text{memenuhi})\\ |2x-3|& =-5+4=-1\\ & (\mathbf{\text{tidak memenuhi}})\\ \text{Selanjutnya}& ,\\ (2x-3)& =\pm 9\\ 2x& =\pm 9+3\\ \text{untuk}& x=9\\ 2x& =9+3\\ x& ={\displaystyle \frac{12}{2}=6}\\ \text{untuk}& x=-9\\ 2x& =-9+3\\ x& ={\displaystyle \frac{-6}{2}=-3}\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Nilai}x\text{yang memenuhi}\\ & \text{persamaan}|x+1|-2|x-3|=5|x-4|\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1\frac{1}{6}\text{atau}1\frac{2}{3}}\\ \\ \text{b}.& {\displaystyle 1\frac{5}{6}\text{atau}2\frac{2}{5}}\\ \\ \text{c}.& {\displaystyle 1\frac{5}{6}\text{atau}2\frac{2}{3}}\\ \\ \text{d}.& {\displaystyle 2\frac{3}{4}\text{atau}4\frac{2}{3}}\\ \\ \text{e}.& {\displaystyle 3\frac{1}{4}\text{atau}4\frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\text{e}\\ & \begin{array}{rl}|x+1|-& 2|x-3|=5|x-4|\\ \text{Perhatik}& \text{an untuk batas sesuai definisi, maka}\\ x=-1& ,x=3,x=4\\ \text{saat}x& \ge 4\\ |x+1|& =x+1,|x-3|=x-3,|x-4|=x-4\\ (x+1)-& 2(x-3)=5(x-4)\\ x-2x-& 5x=-1-6-20\\ -6x& =-27\\ x& ={\displaystyle \frac{27}{6}=4\frac{3}{6}(\text{memenuhi})}\\ \text{saat}3& \le x<4\\ |x+1|& =x+1,|x-3|=x-3,|x-4|=4-x\\ (x+1)-& 2(x-3)=5(4-x)\\ x-2x+& 5x=-1-6+20\\ 4x& =13\\ x& ={\displaystyle \frac{13}{4}=3\frac{1}{4}(\text{memenuhi})}\\ \text{saat}-& 1\le x<3\\ |x+1|& =x+1,|x-3|=3-x,|x-4|=4-x\\ (x+1)-& 2(3-x)=5(4-x)\\ x+2x+& 5x=-1+6+20\\ 8x& =25\\ x& ={\displaystyle \frac{25}{8}=3\frac{1}{8}(\text{tidak memenuhi})}\\ \text{saat}-& 1>x\\ |x+1|& =-x-1,|x-3|=3-x,|x-4|=4-x\\ (-x-1)& -2(3-x)=5(4-x)\\ -x+2x& +5x=1+6+20\\ 6x& =27\\ x& ={\displaystyle \frac{27}{6}=4\frac{3}{6}(\text{tidak memenuhi})}\end{array}\end{array}$

Sumber Referensi

1. Kanginan, M. 2016. Matematika untuk Siswa SMA-MA/SMK-MAK Kelas X. Bandung: Srikandi Empat Widya Utama.
2. 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

$\begin{array}{l}\\ 16.& \text{Penyelesaian dari}|3x-(4x-7)|=6\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \{-1,-13\}\\ \text{b}.& \{-1,13\}\\ \text{c}.& \{1,13\}\\ \text{d}.& \{-13,1\}\\ \text{e}.& \left\{13\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}|3x-(4x-7)|& =6\\ |3x-4x+7|& =6\\ |-x+7|& =6\\ (-x+7)& =\pm 6\\ -x& =\pm 6-7\\ \text{dikalikan}& \text{dengan}-1\\ x& =\pm 6+7\\ x_1& =+6+7=+13\\ x_2& =-6+7=+1\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Penyelesaian}|x-1|=2x+1\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \{-2\}\\ \text{b}.& \{-2,0\}\\ \text{c}.& \{-1\}\\ \text{d}.& \left\{\right\}\\ \text{e}.& \left\{0\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}|x-1|& =2x+1\\ (x-1)& =\pm (2x+1)\\ \text{untuk}& \{\begin{array}{ll}x\ge 1& \text{ maka }(x-1)=2x+1\\ x<1& \text{ maka }(x-1)=-(2x+1)\end{array}\\ & \begin{array}{|cc|}\hline x\ge 1& x<1\\ \begin{array}{rl}x-1& =2x+1\\ x-2x& =1+1\\ -x& =2\\ x& =-2\\ & \end{array}& \begin{array}{rl}x-1& =-(2x+1)\\ x-1& =-2x-1\\ x+2x& =-1+1\\ 3x& =0\\ x& =0\end{array}\\ \text{Tidak memenuhi}& \text{memenuhi}\\ \hline\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Penyelesaian}|3a+1|=2a+9\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \{-2\}\\ \text{b}.& \left\{8\right\}\\ \text{c}.& \{-2,8\}\\ \text{d}.& \left\{\right\}\\ \text{e}.& \text{Semua bilangan riil}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}|3a+1|& =2a+9\\ (3a+1)& =\pm (2a+9)\\ \text{untuk}& \{\begin{array}{ll}a\ge -{\displaystyle \frac{1}{3}}& \text{ maka }(3a+1)=2a+9\\ \\ a<-{\displaystyle \frac{1}{3}}& \text{ maka }(3a+1)=-(2a+9)\end{array}\\ & \begin{array}{|cc|}\hline a\ge {\displaystyle -\frac{1}{3}}& a<-{\displaystyle \frac{1}{3}}\\ \begin{array}{rl}3a+1& =2a+9\\ 3a-2a& =9-1\\ a& =8\\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}3a+1& =-(2a+9)\\ 3a+1& =-2a-9\\ 3a+2a& =-9-1\\ 5a& =-10\\ a& ={\displaystyle \frac{-10}{5}}\\ a& =-2\end{array}\\ \text{memenuhi}& \text{memenuhi}\\ \hline\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Penyelesaian dari}|3x+2|=4x+5\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -3\\ \text{b}.& -1\\ \text{c}.& -3\text{dan}-1\\ \text{d}.& \left\{\right\}\\ \text{e}.& 0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{r}\begin{array}{rl}|3x+2|& =4x+5\\ (3x+2)& =\pm (4x+5)\\ \text{untuk}& \{\begin{array}{ll}x\ge -{\displaystyle \frac{2}{3}}& \text{ maka }3x+2=4x+5\\ \\ x<-{\displaystyle \frac{2}{3}}& \text{ maka }3x+2=-(4x+5)\end{array}\\ & \begin{array}{|cc|}\hline x\ge -{\displaystyle \frac{2}{3}}& x<-{\displaystyle \frac{2}{3}}\\ \begin{array}{rl}3x+2& =4x+5\\ 3x-4x& =5-2\\ -& =3\\ x& =-3\\ & \\ & \end{array}& \begin{array}{rl}3x+2& =-(4x+5)\\ 3x+2& =-4x-5\\ 3x+4x& =-5-2\\ 7x& =-7\\ x& ={\displaystyle \frac{-7}{7}}\\ x& =-1\end{array}\\ \hline\end{array}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Jika}|-3x|+4y^{-1}=6z+4x,\\ & \text{maka nilai}x\text{dinyatakan dalam}y\text{dan}z\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{y}{4}-6z}\\ \text{b}.& {\displaystyle \frac{4}{y}-6z}\\ \text{c}.& 4-{\displaystyle \frac{6}{y}}\\ \text{d}.& {\displaystyle \frac{4}{7y}+\frac{6z}{7}}\\ \text{e}.& {\displaystyle \frac{7}{4y}+6z}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}|-3x|& +4y^{-1}=6z+4x\\ |-3x|& =6z+4x-4y^{-1}\\ -3x& =\pm (6z+4x-4y^{-1})\\ \text{untuk}& :x\ge 0\\ -3x& =6z+4x-4y^{-1}\\ -3x-4x& =6z-4y^{-1}\\ -7x& =6z-4y^{-1}\\ x& ={\displaystyle \frac{6z-4y^{-1}}{-7}}\\ & =\frac{4}{7y}-\frac{6z}{7}\\ untuk& :x<0\\ -3x& =-(6z+4x-4y^{-1})\\ -3x& =-6z-4x+4y^{-1}\\ -3x+4x& =-6z+4y^{-1}\\ x& =4y^{-1}-6z\\ & ={\displaystyle \frac{4}{y}-6z}\end{array}\end{array}$

Contoh Soal Lanjutan 2 Nilai Mutlak

$\begin{array}{l}\\ 11.& \text{Nilai}p\text{yang memenuhi}10-4|4-5p|=26\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 2\text{atau}1{\displaystyle \frac{2}{3}}\\ \text{b}.& 1\text{atau}-{\displaystyle \frac{3}{5}}\\ \\ \text{c}.& 2\text{atau}2{\displaystyle \frac{3}{5}}\\ \text{d}.& -2{\displaystyle \frac{3}{4}\text{atau}1}\\ \text{e}.& -1\text{atau}2{\displaystyle \frac{3}{5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}10-4|4-5p|& =-26\\ -4|4-5p|& =-36\\ |4-5p|& =9\\ (4-5p)& =\pm 9\\ -5p& =-4\pm 9\\ p& ={\displaystyle \frac{-4\pm 9}{-5}}\\ p& =\{\begin{array}{ll}{\displaystyle \frac{-4+9}{-5}}& =-1\\ \text{atau}& \\ {\displaystyle \frac{-4-9}{-5}}& ={\displaystyle \frac{13}{5}=2\frac{3}{5}}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Jika}3<x<5\text{maka penyelesaian untuk}\\ & \sqrt{x^2-6x+9}-\sqrt{x^2-10x+25}=...\\ & \begin{array}{l}\\ \text{a}.& 2x-2\\ \text{b}.& 2\\ \text{c}.& 8-2x\\ \text{d}.& -2\\ \text{e}.& 2x-8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}\sqrt{x^2-6x+9}& -\sqrt{x^2-10x+25}\\ & \sqrt{(x-3{)}^2}-\sqrt{(x-5{)}^2}\\ & =|x-3|-|x-5|\\ & \text{ingat bahwa saat}\\ 3<x<5& \text{maka}\\ & \{\begin{array}{l}|x-3|=(x-3)\\ |x-5|=-(x-5)\end{array},\\ & \text{sehingga}\\ & =|x-3|-|x-5|\\ & =(x-3)-(-(x-5))\\ & =x-3+x-5\\ & =2x-8\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Jika}1<x<5\text{maka penyelesaian untuk}\\ & \sqrt{x^2-2x+1}+\sqrt{x^2-10x+25}=...\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{x^2-2x+1}& +\sqrt{x^2-10x+25}\\ & =\sqrt{(x-1{)}^2}+\sqrt{(x-5{)}^2}\\ & =|x-1|+|x-5|\\ & \text{ingat bahwa saat}\\ 1<x<5& \text{maka}\{\begin{array}{l}|x-1|=(x-1)\\ |x-5|=-(x-5)\end{array},\\ & \text{sehingga}\\ & =|x-1|+|x-5|\\ & =(x-1)+(-(x-5))\\ & =x-1+5-x\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Perhatikanlah ilustrasi grafik di bawah ini}\\ & \begin{array}{}\end{array}\end{array}$

$\begin{array}{l}\\ .& \text{Persamaan yang memenuhi rumus tersebut adalah... .}\\ & \begin{array}{l}\\ \text{a}.& y=|-x-2|\\ \text{b}.& y=-2x-4\\ \text{c}.& y=-|2x-4|\\ \text{d}.& y=|-2x-4|\\ \text{e}.& y=|-2x+4|\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Dengan cara substitusi langsung}\\ & \text{kita akan mendapatkan}\\ & \bullet \text{untuk}x=4\text{menyebabkan nilai}y=4,\\ & \text{maka sampai langkah di sini hanya }\\ & \text{ada 1 persamaan yang memenuhi}\\ & \text{yaitu}:y=|-2x+4|\end{array}\end{array}$

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

$\begin{array}{rl}& \text{berikut gambar grafiknya}\end{array}$