Latihan Soal 7 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{l}\\ 56.& \text{Nilai}x\text{berikut yang tidak memenuhi}\\ & {\displaystyle \frac{x-3}{x^2+2x+1}\le 0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -1\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {\displaystyle \frac{x-3}{x^2+2x+1}\le 0}\\ & {\displaystyle \frac{(x-3)}{(x+1{)}^2}\le 0}\\ & \text{Pembuat nol}\\ & \{\begin{array}{l}x=3,\text{boleh digunakan}\\ x=-1,\text{tetapi}x\ne -1,\end{array}\\ & \text{sehingga}-1\text{tidak digunakan}\\ & \begin{array}{c}\\ & & & & & & & & & & \\ & -& -& -& -& -& & +& +& & \\ & & -1& & & & 3& & & & \\ & & & & & & & & & & \end{array}& \end{array}\end{array}$.

$\begin{array}{l}\\ 57.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x-5}\le x\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 1<x<0\\ \\ \text{b}.& x<1\text{atau}x\ge 5\\ \\ \text{c}.& {\displaystyle \frac{5}{6}\le x\le 1\text{atau}x\ge 5}\\ \\ \text{d}.& {\displaystyle \frac{5}{6}\le x<1\text{atau}5<x<6}\\ \\ \text{e}.& x\ge 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{6x-5}& \le x\\ 1.\text{Kuadrat}& \text{kan}\\ 6x-5& \le x^2\\ -x^2+6x-5& \le 0\\ x^2-6x+5& \ge 0\\ (x-1)(x-5)& \ge 0\\ x\le 1\text{atau}& x\ge 5\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x-5& \ge 0\\ 6x& \ge 5\\ x& \ge {\displaystyle \frac{5}{6}}\end{array}\end{array}$

$\begin{array}{l}\\ 58.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x+6}>6\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x>7\\ \text{b}.& x\ge 7\\ \text{c}.& x<7\\ \text{d}.& x>1\\ \text{e}.& x\ge 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\sqrt{6x+6}& >6\\ 1.\text{Kuadrat}& \text{kan}\\ 6x+6& >36\\ x+1& >6\\ x& >7\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x+6& \ge 0\\ 6x& \ge 6\\ x& \ge {\displaystyle \frac{6}{6}}\\ x& \ge 1\end{array}\end{array}$

$\begin{array}{l}\\ 59.& \text{Penyelesaian pertidaksamaan}\\ & x+2>{\displaystyle \sqrt{10-x^2}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\le x\le \sqrt{10}\\ \text{b}.& 1<x\le \sqrt{10}\\ \text{c}.& -3<x\le \sqrt{10}\\ \text{d}.& -\sqrt{10}\le x\le \sqrt{10}\\ \text{e}.& x<-3\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}x+2& >{\displaystyle \sqrt{10-x^2}}\\ 1.\text{Kuadrat}& \text{kan}\\ x^2+4x+4& >10-x^2\\ 2x^2+4x& +4-10>0\\ 2x^2+4x-& 6>0\\ x^2+2x-& 3>0\\ (x+3)& (x-1)>0\\ x<-3& \text{atau}x>1\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 10-x^2& \ge 0\\ x^2-10& \le 0\\ (x-\sqrt{10})& (x+\sqrt{10})\le 0\\ -\sqrt{10}\le x& \le \sqrt{10}\end{array}\end{array}$

$\begin{array}{l}\\ 60.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{3x+7}\ge x-1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -1<x<6\\ \text{b}.& -1\le x<6\\ \text{c}.& x\ge -{\displaystyle \frac{7}{3}}\\ \text{d}.& -{\displaystyle \frac{7}{3}\le x\le 6}\\ \text{e}.& -{\displaystyle \frac{7}{3}\le x\le 1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{3x+7}& \ge x-1\\ 1.\text{Kuadrat}& \text{kan}\\ 3x+7& \ge x^2-2x+1\\ -x^2+3x& +2x+7-1\ge 0\\ -x^2+5x+& 6\ge 0\\ x^2-5x-& 6\le 0\\ (x+1)& (x-6)\le 0\\ -1\le x& \le 6\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 3x+7& \ge 0\\ 3x& \ge -7\\ x& \ge -{\displaystyle \frac{7}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 61.& \text{Nilai}x\text{yang memenuhi}\\ & \sqrt{2x-8}<\sqrt{x+5}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x\ge -5\\ \text{b}.& x<-13\text{atau}x\ge -4\\ \text{c}.& x<13\\ \text{d}.& 4\le x<13\\ \text{e}.& -5\le x\le 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{2x-8}& <\sqrt{x+5}\\ (1)\text{kuadratkan}& \\ 2x-8& <x+5\\ x& <13\\ (2)2x-8\ge 0& \\ x& \ge 4\\ (3)x+5\ge 0& \\ x& \ge -5\\ \text{perhatikan}& \text{lah garis bilangannya berikut}\end{array}\end{array}$.

$\begin{array}{l}\\ 62.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x-4}<\sqrt{2x+8}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -4<x\le {\displaystyle \frac{2}{3}}\\ \text{b}.& -4<x<3\\ \text{c}.& {\displaystyle \frac{2}{3}\le x<3}\\ \text{d}.& 2<x\le 4\\ \text{e}.& -4\le x\le 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \sqrt{6x-4}}& <\sqrt{2x+8}\\ 1.\text{Kuadrat}& \text{kan}\\ 6x-4& <2x+8\\ 6x-2x& <8+4\\ 4x& <12\\ x& <3\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x-4& \ge 0\\ 6x& \ge 4\\ x& \ge {\displaystyle \frac{2}{3}}\\ 3.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 2x+8& \ge 0\\ 2x& \ge -8\\ x& \ge -4\end{array}\end{array}$

$\begin{array}{l}\\ 63.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{x+3}>\sqrt{12-2x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 3<x\le 6\\ \text{b}.& -3<x\le 6\\ \text{c}.& -6<x\le 3\\ \text{d}.& -6<x\le -3\\ \text{e}.& x<3\text{atau}x>6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle {\displaystyle \sqrt{x+3}}}& >\sqrt{12-2x}\\ 1.\text{Kuadrat}& \text{kan}\\ x+3& >12-2x\\ x+2x& >12-3\\ 3x& >9\\ x& >3\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ x+3& \ge 0\\ x& \ge -3\\ 3.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 12-2x& \ge 0\\ 2x-12& \le 0\\ 2x& \le 12\\ x& \le 6\end{array}\end{array}$

$\begin{array}{l}\\ 64.& (\mathbf{\text{SBMPTN 2013 Mat Das}})\\ & \text{Jika}1<m<2,\text{maka semua nilai}\\ & x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2+4x}{-x^2+3x-3m}>0\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x>-3\\ \text{b}.& x<-4\\ \text{c}.& -4<x<0\\ \text{d}.& x<-4\text{atau}x>0\\ \text{e}.& x<-3\text{atau}x>-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}1.& \text{Diketahui bahwa}:{\displaystyle \frac{x^2+4x}{-x^2+3x-3m}>0}\\ & \text{dengan kondisi}1<m<2\\ & \text{Perhatikanlah penyebutnya yang}\\ & \text{mengandung bilangan}m\text{yang terletak}\\ & \text{pada interval}:1<m<2.\\ 2.& \text{Kita cek kondisi penyebutnya dengan}\\ & \text{menentukan}Diskriminan(D)-\text{nya}\\ & \text{yaitu}:\\ & a{x}^2+bx+c\{\begin{array}{l}a>0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit positif}\\ a<0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit negatif}\end{array}\\ & \text{Karena penyebutnya}:-x^2+3x-3m,\\ & \text{dengan}a=-1,b=3,\mathrm{\&}c=-3m,\text{maka}\\ & D=3^2-4(-1)(-3m)=9-12m\\ 3.& \text{Karena nilai}m\text{berada pada}1<m<2\\ & \text{maka}\\ & 1<m<2\\ & \iff 12<12m<24\\ & \iff -12>-12m>-24\\ & \iff 9-12>9-12m>-13\\ & \iff -3>9-12m>-13\\ & \iff -13<9-12m<-3\\ & \text{Ini berarti nilai}D\text{negatif, sehingga}\\ & \text{berakibat penyebut berupa}-x^2+3x-3m\\ & \text{adalah wilayah}definitnegatif\\ 4.& \text{Selanjutnya pemfaktoran pertidaksamaan}\\ & \bullet \text{semula}\\ & {\displaystyle \frac{x(x+4)}{\underset{definitnegatif}{\underbrace{-x^2+3x-3m}}}>0\iff \frac{x(x+4)}{-}>0}\\ & \bullet \text{akan berubah menjadi}\\ & x(x+4)<0\\ & \text{pembuat nol-nya adalah}:x(x+4)=0\\ & \text{maka}x=-4\text{atau}x=0,\text{sehingga}\\ & \text{interval nilai}x-\text{nya}:-4<x<0\end{array}\end{array}$

$\begin{array}{l}\\ 65.& \text{Jika}1<a<2,\text{maka semua nilai}\\ & x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{-x^2+2ax-6}{x^2+3x}\le 0\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x<-3\text{atau}x>0\\ \text{b}.& x<-3\text{atau}x\ge -2\\ \text{c}.& x\le -2\text{atau}x\ge 2\\ \text{d}.& -3<x<0\\ \text{e}.& -2\le x<0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}1.& \text{Diketahui bahwa}:{\displaystyle \frac{-x^2+2ax-6}{x^2+3x}\le 0}\\ & \text{untuk membedakan}a\text{pada persamaan}\\ & \text{kuadrat dengan}a\text{di atas, selanjutnya}\\ & \text{kita menuliskan}a\text{di atas dengan}:m\\ & \text{karena}1<a<2\text{diubah}:1<m<2\\ & \text{Perhatikanlah pembilang yang}\\ & \text{mengandung bilangan}m\text{yang terletak}\\ & \text{pada interval}:1<m<2.\\ 2.& \text{Kita cek kondisi pembilangnya dengan}\\ & \text{menentukan}Diskriminan(D)-\text{nya}\\ & \text{yaitu}:\\ & a{x}^2+bx+c\{\begin{array}{l}a>0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit positif}\\ a<0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit negatif}\end{array}\\ & \text{Karena pebilangnya}:-x^2+2mx-6,\\ & \text{dengan}a=-1,b=2m,\mathrm{\&}c=-6,\text{maka}\\ & D=(2m{)}^2-4(-1)(-6)=4m^2-24\\ 3.& \text{Karena nilai}m\text{berada pada}1<m<2\\ & \text{maka}\\ & 1<m<2\\ & \iff 1^2<m^2<2^2\iff 1<m^2<4\\ & \iff 4<4m^2<16\\ & \iff 4-24<4m^2-24<16-24\\ & \iff -20<4m^2-24<-8\\ & \text{Ini berarti nilai}D\text{negatif, sehingga}\\ & \text{berakibat pembilangnya berupa}-x^2+2mx-6\\ & \text{adalah wilayah}definitnegatif\\ 4.& \text{Selanjutnya pemfaktoran pertidaksamaan}\\ & \bullet \text{semula}\\ & {\displaystyle \frac{\stackrel{definitnegatif}{\overbrace{-x^2+2mx-6}}}{x(x+3)}\le 0\iff \frac{-}{x(x+3)}\le 0}\\ & \bullet \text{akan berubah menjadi}\\ & x(x+3)>0\\ & \text{pembuat nol-nya adalah}:x(x+3)=0\\ & \text{maka}x=-3\text{atau}x=0,\text{sehingga}\\ & \text{interval nilai}x-\text{nya}:x<-3\text{atau}x>0\end{array}\end{array}$.

Latihan Soal 6 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 46.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{2}{x+1}\le \left|x\right|\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& \{x|x\le -2\text{atau}x\ge 1\}\\ \text{b}.& \{x|x\le -2\text{atau}0\le x\le 1\}\\ \text{c}.& \{x|x\ge 1\}\\ \text{d}.& \{x|x<-1\text{atau}x\ge 1\}\\ \text{e}.& \{x|-1<x\le 1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \left|x\right|\ge {\displaystyle \frac{2}{x+1}\text{berakibat}}\\ & {\displaystyle \frac{-2}{x+1}\ge x\text{atau}x\ge {\displaystyle \frac{2}{x+1}}}\\ & \bullet \text{bagian}1\\ & x\le {\displaystyle \frac{-2}{x+1}(\mathbf{\text{tidak boleh kali silang}})}\\ & x+{\displaystyle \frac{2}{x+1}\le 0}\\ & {\displaystyle \frac{x(x+1)+2}{x+1}\le 0}\\ & {\displaystyle \frac{x^2+x+2}{x+1}\le 0\iff {\displaystyle \frac{\text{Definit positif}}{x+1}\le 0}}\\ & \begin{array}{rl}& {\text{HP}}_1=\{x|x<-1,x\in \mathbb{R}\}\\ & \bullet \text{bagian}2\\ & x\ge {\displaystyle \frac{2}{x+3}}\\ & x-{\displaystyle \frac{2}{x+1}\ge 0}\\ & {\displaystyle \frac{x(x+1)-2}{x+1}\ge 0}\\ & {\displaystyle \frac{x^2+x-2}{x+1}\ge 0}\\ & {\displaystyle \frac{(x+2)(x-1)}{x+1}\ge 0}\\ & {\text{HP}}_2=\{x|-2\le x<-1\text{atau}x\ge 1,x\in \mathbb{R}\}\\ & \text{HP}={\text{HP}}_1+{\text{HP}}_2=\{x|x<-1\text{atau}x\ge 1,x\in \mathbb{R}\}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 47.& \text{Diketahui pertidaksamaan}{\displaystyle \frac{x+10}{x-9}\le 0}\\ & \text{dan diberikan beberapa nilai berikut}\\ & (\text{i})x=-6(\text{iii})x=-14\\ & (\text{ii})x=-10(\text{iv})x=-18\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \text{di atas adalah ditunjukkan oleh}....\\ & \begin{array}{l}\\ \text{a}.& (\text{i})\text{dan}(\text{ii})\\ \text{b}.& (\text{i})\text{dan}(\text{iii})\\ \text{c}.& (\text{ii})\text{dan}(\text{iii})\\ \text{d}.& (\text{ii})\text{dan}(\text{iv})\\ \text{e}.& (\text{iii})\text{dan}(\text{iv})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{x+10}{x-9}}& \le 0\\ \text{HP}=& \{x|-10\le x<9,x\in \mathbb{R}\}\end{array}\end{array}$.

$\begin{array}{l}\\ 48.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\text{ adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 2\le x<6\\ \text{b}.& 2\le x<3\\ \text{c}.& 2<x<3\text{atau}x>6\\ \text{d}.& x<3\text{atau}3<x<6\\ \text{e}.& x<2\text{atau}x>3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {\displaystyle \frac{6}{x-3}<\frac{8}{x-2}}\\ & \iff {\displaystyle \frac{6}{x-3}-\frac{8}{x-2}<0}\\ & \iff {\displaystyle \frac{6(x-2)-8(x-3)}{(x-3)(x-2)}<0}\\ & \iff {\displaystyle \frac{6x-8x-12+24}{(x-2)(x-3)}<0}\\ & \iff {\displaystyle \frac{-2x+12}{(x-2)(x-3)}<0}\\ & \iff {\displaystyle \frac{2x-12}{(x-2)(x-3)}>0}\\ & \iff {\displaystyle \frac{2(x-6)}{(x-2)(x-3)}>0}\\ & \text{HP}=\{x|2<x<3\text{atau}x>6,x\in \mathbb{R}\}\end{array}\end{array}$.

$\begin{array}{l}\\ 49.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \frac{x^2-81}{x^2}\ge 0\text{ adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x\le -9\text{atau}x\ge 9\\ \text{b}.& -9\le x<0\text{atau}x\ge 9\\ \text{c}.& -9\le x<0\text{atau}0<x\le 9\\ \text{d}.& -9<x\le 9\\ \text{e}.& x\in \mathbb{R}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{x^2-81}{x^2}\ge 0}\\ & {\displaystyle \frac{(x+9)(x-9)}{x^2}\ge 0}\\ & \text{HP}=\{x|x\le -9\text{atau}x\ge 9,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 50.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2-4}{x+2}>0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x>2\\ \text{b}.& -2\le x<2\\ \text{c}.& x<-2\text{atau}x>2\\ \text{d}.& x<-2\text{atau}-2<x<2\\ \text{e}.& x\ge -2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{x^2-4}{x+2}>0}\\ & {\displaystyle \frac{(x+2)(x-2)}{(x+2)}>0}\\ & (x-2)>0\\ & x>2\end{array}\end{array}$.

$\begin{array}{l}\\ 51.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2+x-30}{2x^2+13x-45}<0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& \{x|-9<x<5,x\in \mathbb{R}\}\\ \text{b}.& \{x|-6<x<5,x\in \mathbb{R}\}\\ \text{c}.& \{x|-9<x<-6\text{atau}x<5,x\in \mathbb{R}\}\\ \text{d}.& \{x|-9<x<-6\text{atau}{\displaystyle \frac{5}{2}<x<5,x\in \mathbb{R}}\}\\ \text{e}.& \{x|x<-9\text{atau}-6<x<{\displaystyle \frac{5}{2}\text{atau}x<5,x\in \mathbb{R}}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2+x-30}{2x^2+13x-45}}& <0\\ {\displaystyle \frac{(x+6)(x-5)}{(x+9)(2x-5)}}& <0\\ \text{Cukup jelas}& \end{array}\end{array}$.

$\begin{array}{l}\\ 52.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{2x+6}{x-4}\le 1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -10<x<4\\ \text{b}.& -10\le x<4\\ \text{c}.& -4<x\le 10\\ \text{d}.& x\le -10\text{atau}x\ge 4\\ \text{e}.& x<-10\text{atau}x\ge 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{2x+6}{x-4}}& \le 1\\ {\displaystyle \frac{2x+6}{x-4}-1}& \le 0\\ {\displaystyle \frac{2x+6-(x-4)}{x-4}}& \le 0\\ {\displaystyle \frac{x+10}{x-4}}& \le 0\end{array}\end{array}$

$\begin{array}{l}\\ 53.& \text{Nilai}x\text{yang memenuhi}{\displaystyle \frac{x^2-x}{x+3}\ge 1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x<-3\text{atau}-1\le x\le 3\\ \text{b}.& -3<x\le -1\text{atau}x\ge 3\\ \text{c}.& -3\le x\le 3\\ \text{d}.& -3\le x\le -1\text{atau}x\ge 3\\ \text{e}.& -3\le x\le -1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2-x}{x+3}}& \ge 1\\ {\displaystyle \frac{x^2-x}{x+3}}& -1\ge 0\\ {\displaystyle \frac{x^2-x-(x+3)}{x+3}}& \ge 0\\ {\displaystyle \frac{x^2-2x-3}{x+3}}& \ge 0\\ {\displaystyle \frac{(x-3)(x+1)}{x+3}}& \ge 0\end{array}\end{array}$

$\begin{array}{l}\\ 54.& \text{Nilai}x\text{yang memenuhi}\\ & x+2+{\displaystyle \frac{1}{x+4}>0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x<-4\text{atau}x\ge -3\\ \text{b}.& x<-4\text{atau}x>-3\\ \text{c}.& -4\le x\le -3\\ \text{d}.& x>-4\\ \text{e}.& -4\le x\le -3\text{atau}x>-3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}x+2+{\displaystyle \frac{1}{x+4}}& >0\\ {\displaystyle \frac{(x+2)(x+4)+1}{(x+4)}}& >0\\ {\displaystyle \frac{x^2+6x+8+1}{x+4}}& >0\\ {\displaystyle \frac{x^2+6x+9}{x+4}}& >0\\ {\displaystyle \frac{(x+3{)}^2}{(x+4)}}& >0\\ x& >-4\end{array}\end{array}$

$\begin{array}{l}\\ 55.& \text{Nilai}x\text{yang memenuhi}\\ & x+3<{\displaystyle \frac{x^2+6x+11}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-3{\displaystyle \frac{2}{3}\text{atau}x>0,x\in \mathbb{R}}\}\\ \text{b}.& \{x|0\le x\le 11,x\in \mathbb{R}\}\\ \text{c}.& \{x|x<-11\text{atau}x>0,x\in \mathbb{R}\}\\ \text{d}.& \{x|x<0\text{atau}x>11,x\in \mathbb{R}\}\\ \text{e}.& \{x|x\le 11\text{atau}x>0,x\in \mathbb{R}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}x+3<{\displaystyle \frac{x^2+6x+11}{x}}& \\ x+3-{\displaystyle \frac{x^2+6x+11}{x}}& <0\\ {\displaystyle \frac{x(x+3)-(x^2+6x+11)}{x}}& <0\\ {\displaystyle \frac{x^2+3x-x^2-6x-11}{x}}& <0\\ {\displaystyle \frac{-3x-11}{x}}& <0\\ {\displaystyle \frac{3x+11}{x}}& >0\end{array}\end{array}$.

Latihan Soal 5 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{l}\\ 36.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |3-\left|x\right||<10\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-14\text{atau}x>12\\ \text{b}.& x<-13\text{atau}x>13\\ \text{c}.& x<-12\text{atau}x>10\\ \text{d}.& 0<x<10\\ \text{e}.& -13<0<13\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& |3-\left|x\right||<10\\ & -10<3-\left|x\right|<10\\ & -13<-\left|x\right|<7\\ & -7<\left|x\right|\le 13,(\text{ingat harga}\left|x\right|\ge 0)\\ & 0\le \left|x\right|<13\\ & \text{selanjutnya},\\ & \left|x\right|<13\\ & -13<x<13\\ & \text{HP}=\{x|-13<x<13,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \text{(UM UGM 05)}\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |x^2-3|<2x\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -1<x<3\\ \text{b}.& -3<x<1\\ \text{c}.& 1<x<3\\ \text{d}.& -3<x<-1\text{atau}1<x<3\\ \text{e}.& x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}|x^2-3|& <2x\\ -2x<(x^2-3)& <2x\\ \text{dipartisi men}& \text{jadi dua bagian}\\ \bullet \text{pertama}& \\ (x^2-3)& >-2x\\ x^2+2x-3& >0\\ (x+3)(x-1)& >0\\ x<-3\text{atau}& x>1\\ \bullet \text{kedua}& \\ (x^2-3)& <2x\\ x^2-2x-3& <0\\ (x-3)(x+1)& <0\\ -1<x<3& \\ \text{ambil yang}& \text{memenuhi keduanya}\\ \text{berupa iris}& \text{an}\\ \text{HP}=& \{1<x<3,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 38.& (\text{SPMB 05})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {|x-2|}^2<4|x-2|+12\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \{x\in \mathbb{R}|2\le x\le 8\}\\ \text{b}.& \{x\in \mathbb{R}|4<x<8\}\\ \text{c}.& \{x\in \mathbb{R}|-4<x<8\}\\ \text{d}.& \{x\in \mathbb{R}|-2<x<4\}\\ \text{e}.& \{x\in \mathbb{R}|2<x<4\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{misalkan}p& =|x-2|,\text{selanjutnya}\\ {|x-2|}^2<& 4|x-2|+12\\ p^2<& 4p+12\\ p^2-& 4p-12<0\\ (p-6)& (p+2)<0\\ -2<p& <6,\text{atau jika dikembalikan}\\ -2<& |x-2|<6,\\ & \text{ingat, nilanya tidak negatif}\\ 0\le & |x-2|<6\\ -6<& x-2<6\\ -4<& x<8\\ \text{HP}=& \{-4<x<8,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Diketahui grafik fungsi}f(x)=m{x}^2-2mx+m\\ & \text{berada di atas grafik fungsi}\\ & g(x)=2x^2-3,\text{maka nilai}m\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& m>2& & & \text{d}.& -6<m<2\\ \text{b}.& m>6& \text{c}.& 2<m<6& \text{e}.& m<-6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(x)& =g(x)\\ m{x}^2-2mx+m& =2x^2-3\\ m{x}^2-2x^2-2mx+m+3& =0\\ \text{Supaya grafik}f(x)\text{berada }& \text{di atasnya},\\ \text{maka}D=B^2-4AC& <0\\ (m-2)x^2-2mx+(m+3)& =0\{\begin{array}{ll}A& =m-2\\ B& =-2m\\ C& =m+3\end{array}\\ B^2-4AC& <0\\ (-2m{)}^2-4(m-2)(m+3)& <0\\ 4m^2-4(m^2+m-6)& <0\\ 4m^2-4m^2-4m+24& <0\\ -4m+24& <0\\ m-6& >0\\ m& >6\end{array}\end{array}$.

$\begin{array}{l}\\ 40.& \text{Jika}3<x<5\text{maka penyelesaian untuk}\\ & \sqrt{x^2-6x+9}-\sqrt{x^2-10x+25}=...\\ & \begin{array}{l}\\ \text{a}.& 2x-2& & & \text{d}.& -2\\ \\ \text{b}.& 2& \text{c}.& 8-2x& \text{e}.& 2x-8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \sqrt{x^2-6x+9}-\sqrt{x^2-10x+25}\\ & =\sqrt{(x-3{)}^2}-\sqrt{(x-5{)}^2}\\ & =|x-3|-|x-5|\\ & \text{ingat bahwa saat}3<x<5\\ & \text{maka}\{\begin{array}{l}|x-3|=(x-3)\\ |x-5|=-(x-5)\end{array},\\ & \text{sehingga}\\ & =|x-3|-|x-5|=(x-3)-(-(x-5))\\ & =x-3+x-5\\ & =2x-8\end{array}\end{array}$.

$\begin{array}{l}\\ 41.& \text{Jika}1<x<5\text{maka penyelesaian untuk}\\ & \sqrt{x^2-2x+1}+\sqrt{x^2-10x+25}=...\\ & \begin{array}{l}\\ \text{a}.& 2& & & \text{d}.& 5\\ \\ \text{b}.& 3& \text{c}.& 4& \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \sqrt{x^2-2x+1}+\sqrt{x^2-10x+25}\\ & =\sqrt{(x-1{)}^2}+\sqrt{(x-5{)}^2}\\ & =|x-1|+|x-5|\\ & \text{ingat bahwa saat}1<x<5\\ & \text{maka}\{\begin{array}{l}|x-1|=(x-1)\\ |x-5|=-(x-5)\end{array},\\ & \text{sehingga}\\ & =|x-1|+|x-5|=(x-1)+(-(x-5))\\ & =x-1+5-x\\ & =4\end{array}\end{array}$.

$\begin{array}{l}\\ 42.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{5}{4x-3}}\right|\le 1\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\le x\le \frac{3}{4}\text{atau}x\ge 2}\\ \text{b}.& x\le -{\displaystyle \frac{1}{2}\text{atau}{\displaystyle \frac{3}{4}<x\le 2}}\\ \text{c}.& -{\displaystyle \frac{1}{2}\le x\le 2,x\ne \frac{3}{4}}\\ \text{d}.& x\le -{\displaystyle \frac{1}{2}\text{atau}x>\frac{3}{4}}\\ \text{e}.& x\le -{\displaystyle \frac{1}{2}\text{atau}x\ge 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{5}{4x-3}}\right|& \le 1\\ -1\le {\displaystyle \frac{5}{4x-3}}& \le 1,\mathbf{\text{jika dibalik}}\\ -1\ge {\displaystyle \frac{4x-3}{5}}& \ge 1,\mathbf{\text{bentuk ini tidak}}\\ \mathbf{\text{dibolehkan}}& \mathbf{\text{maka perlu diubah menjadi}}\\ -1\ge {\displaystyle \frac{4x-3}{5}\text{atau}}& {\displaystyle \frac{4x-3}{5}\ge 1,\text{selanjutnya}}\\ \bullet \text{bagian}& 1\\ -1& \ge {\displaystyle \frac{4x-3}{5}\iff \frac{4x-3}{5}\le -1}\\ 4x-3& \le -5\\ 4x& \le -2\\ x& \le -{\displaystyle \frac{1}{2}}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{4x-3}{5}}& \ge 1\\ 4x-3& \ge 5\\ 4x& \ge 8\\ x& \ge 2\end{array}\end{array}$

$\begin{array}{l}\\ 43.& (\text{UMPTN 95})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{2}{2x-1}}\right|>1\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x>2\\ \text{b}.& x<2\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{c}.& x<-1\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{d}.& -1<x<2\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{e}.& x<-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{semua opsi bukan jawaban}}\\ & \mathbf{\text{Berikut pembahasannya}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{2}{2x-1}}\right|& >1\\ -1>{\displaystyle \frac{2}{2x-1}}& \text{atau}{\displaystyle \frac{2}{2x-1}>1,\mathbf{\text{dibalik}}}\\ -1<{\displaystyle \frac{2x-1}{2}}& \text{atau}{\displaystyle \frac{2x-1}{2}<1}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{2x-1}{2}}& >-1\\ 2x-1& >-2\\ 2x& >-1\\ x& >-{\displaystyle \frac{1}{2}}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{2x-1}{2}}& <1\\ 2x-1& <2\\ 2x& <3\\ x& <{\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 44.& (\text{UMPTN 00})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{2x+7}{x-1}}\right|\ge 1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2\le x\le 8\\ \text{b}.& x\le -8\text{atau}x\ge -2\\ \text{c}.& -8\le x<1\text{atau}x>1\\ \text{d}.& -2\le x<1\text{atau}1<x\le 8\\ \text{e}.& x\le -8\text{atau}-2\le x<1\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{2x+7}{x-1}}\right|& \ge 1\\ -1\ge {\displaystyle \frac{2x+7}{x-1}}& \text{atau}{\displaystyle \frac{2x+7}{x-1}\ge 1}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{2x+7}{x-1}}& \le -1(\mathbf{\text{tidak boleh kali silang}})\\ {\displaystyle \frac{2x+7}{x-1}}& +1\le 0\\ & {\displaystyle \frac{2x+7+(x-1)}{x-1}\le 0}\\ {\displaystyle \frac{3x+6}{x-1}}& \le 0\\ {\text{HP}}_1=& \{x|-2\le x<1,x\in \mathbb{R}\}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{2x+7}{x-1}}& \ge 1\\ {\displaystyle \frac{2x+7}{x-1}}& -1\ge 0\\ & {\displaystyle \frac{2x+7-(x-1)}{x-1}\ge 0}\\ {\displaystyle \frac{x+8}{x-1}}& \ge 0\\ {\text{HP}}_2=& \{x|x\le -8\text{atau}x>1,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x\le -8\text{atau}-2\le x<1\text{atau}x>1,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 45.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{x-2}{x+3}}\right|\le 2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -8\le x<-3\\ \text{b}.& -8\le x<-1\\ \text{c}.& -4\le x<-3\\ \text{d}.& x\le -8\text{atau}x\ge -{\displaystyle \frac{4}{3}}\\ \text{e}.& x\le -4\text{atau}x\ge 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{x-2}{x+3}}\right|& \le 2\\ -2\le {\displaystyle \frac{x-2}{x+3}}& \text{atau}{\displaystyle \frac{x+2}{x+3}\le 2}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{x-2}{x+3}}& \ge -2(\mathbf{\text{tidak boleh kali silang}})\\ {\displaystyle \frac{x-2}{x+3}}& +2\ge 0\\ & {\displaystyle \frac{x-2+2(x+3)}{x+3}\ge 0}\\ {\displaystyle \frac{3x+4}{x+3}}& \ge 0\\ {\text{HP}}_1=& \{x|x<-3\text{atau}x\ge -{\displaystyle \frac{4}{3},x\in \mathbb{R}}\}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{x-2}{x+3}}& \le 2\\ {\displaystyle \frac{x-2}{x+3}}& -2\le 0\\ & {\displaystyle \frac{x-2-2(x+3)}{x+3}\le 0}\\ {\displaystyle \frac{-x-8}{x+3}}& \le 0,\mathbf{\text{koefisien textit{x} negatif}}\\ {\displaystyle \frac{x+8}{x+3}}& \ge 0\\ {\text{HP}}_2=& \{x|x\le -8\text{atau}x>-3,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x\le -8\text{atau}x\ge -{\displaystyle \frac{4}{3},x\in \mathbb{R}}\}\end{array}\end{array}$

Latihan Soal 4 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 26.& \text{Nilai}x\text{yang memenuhi}\\ & (x+3)(x-1)\ge (x-1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\le x\le 3& \\ \text{b}.& x\le -2\text{atau}x\ge 1\\ \text{c}.& -3\le x\le -1\\ \text{d}.& -2\ge x\text{atau}x\ge 3\\ \text{e}.& -1\ge x\text{atau}x\ge 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}(x+3)(x-1)& \ge (x-1)\\ (x+3)(x-1)-(x-1)& \ge 0\\ (x-1)((x+3)-1)& \ge 0\\ (x-1)(x+2)& \ge 0\end{array}\\ & \text{Sehingga solusinya adalah:}\\ & x\le -2\text{atau}x\ge 1\end{array}$.

$\begin{array}{l}\\ 27.& \text{Himpunan penyelesaian pertidaksamaan}\\ & |x+3|<2|x-4|\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-{\displaystyle \frac{5}{3}}\}\\ \text{b}.& \left\{x|{\displaystyle \frac{5}{3}<x<-11}\right\}\\ \text{c}.& \{x|x\ge -11\}\\ \text{d}.& \{x|x<{\displaystyle \frac{5}{3}}\}\cup \{x|x>11\}\\ \text{e}.& \{x|x>-{\displaystyle \frac{5}{3}}\}\cup \{x|x<-11\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& |x+3|<2|x-4|\\ & {(x+3)}^2<2^2{(x-4)}^2\\ & \text{dikuadratkan masing-masing ruas}\\ & x^2+6x+9<4(x^2-8x+16)\\ & x^2-4x^2+6x+32x+9-64<0\\ & -3x^2+38x-55<0\\ & 3x^2-38x+55>0\\ & (3x-5)(x-11)>0\\ \\ & \text{Berikut untuk}\text{garis bilangannya}\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& \text{Himpunan penyelesaian pertidaksamaan}\\ & |x^2+5x|\le 6\text{adalah... .}\\ & \text{a}.\{x|-6\le x\le 1\}\\ & \text{b}.\{x|-3\le x\le -2\}\\ & \text{c}.\{x|-6\le x\le -3\text{atau}-2\le x\le 1\}\\ & \text{d}.\{x|-6\le x\le -5\text{atau}-2\le x\le 0\}\\ & \text{e}.\{x|-5\le x\le -3\text{atau}-2\le x\le 0\}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{r}\text{Diketahui bahwa}\\ & \\ |x^2+5x|& \le 6\\ -6\le x^2+5x& \le 6\\ & \end{array}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}& -6\le x^2+5x\\ & x^2+5x+6\ge 0\\ & (x+3)(x+2)\ge 0\end{array}& \begin{array}{rl}& x^2+5x\le 6\\ & x^2+5x-6\le 0\\ & (x+6)(x-1)\le 0\end{array}\\ \mathbf{\text{Lihat Gambar 1}}& \mathbf{\text{Lihat Gambar 2}}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Nilai}x\text{yang memenuhi}\left|{\displaystyle \frac{1}{2}x+6}\right|\ge 9\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -12<x<6\\ \text{b}.& -30\le x\le 6\\ \text{c}.& x\ge 6\text{atau}x\le -30\\ \text{d}.& x<6\text{atau}x<-30\\ \text{e}.& x\le 6\text{atau}x\ge -30\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \left|{\displaystyle \frac{1}{2}x+6}\right|\ge 9\\ & {\displaystyle \frac{1}{2}x+6\le -9\text{atau}{\displaystyle \frac{1}{2}x+6\ge 9}}\\ & {\displaystyle \frac{1}{2}x\le -9-6\text{atau}{\displaystyle \frac{1}{2}x\ge 9-6}}\\ & {\displaystyle \frac{1}{2}x\le -15\text{atau}{\displaystyle \frac{1}{2}x\ge 3}}\\ & x\le -30\text{atau}x\ge 6\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & 3|x+1|\le |x-2|\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{4}\le x\le -\frac{1}{4}}\\ \text{b}.& -{\displaystyle \frac{5}{2}\le x\le \frac{5}{2}}\\ \text{c}.& x\le {\displaystyle \frac{1}{4}\text{atau}x\ge \frac{5}{2}}\\ \text{d}.& x\le -{\displaystyle \frac{5}{2}\text{atau}x\ge \frac{1}{4}}\\ \text{e}.& x\le -{\displaystyle \frac{5}{2}\text{atau}x\ge -\frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& 3|x+1|\le |x-2|\\ & {\left(3|x+1|\right)}^2\le {\left(|x-2|\right)}^2\\ & {(3x+3)}^2\le {(x-2)}^2\\ & (3x+3+(x-2))(3x+3-(x-2))\le 0\\ & (4x+1)(2x+5)\le 0\\ & \text{HP}=\{x|-{\displaystyle \frac{5}{2}\le x\le -\frac{1}{4},x\in \mathbb{R}}\}\end{array}\end{array}$.

$\begin{array}{l}\\ 31.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |x-3|<3\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<3\\ \text{b}.& -3<x<3\\ \text{c}.& x<-3\text{atau}x<3\\ \text{d}.& x>0\text{atau}x<6\\ \text{e}.& x<0\text{atau}x<6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& |x-3|<3\\ & -3<(x-3)<3\\ & -3+3<x<3+3\\ & 0<x<6\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |x+4|>8\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x>-8\\ \text{b}.& x<4\text{atau}x>12\\ \text{c}.& x>4\text{atau}x>-12\\ \text{d}.& x<-4\text{atau}x<6\\ \text{e}.& x>4\text{atau}x<-12\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& |x+4|>8\\ & (x+4)<-8\text{atau}(x+4)>8\\ & x<-12\text{atau}x>4\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Himpunan penyelesaian dari pertidaksamaan}\\ & \left|{\displaystyle \frac{x+1}{2}}\right|>\left|{\displaystyle \frac{x-2}{3}}\right|\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \text{HP}=\{x|-7<x<{\displaystyle \frac{1}{5},x\in \mathbb{R}}\}\\ \text{b}.& \text{HP}=\{x|x<-7\text{atau}x>{\displaystyle \frac{1}{5},x\in \mathbb{R}}\}\\ \text{c}.& \text{HP}=\{x|x>-7,x\in \mathbb{R}\}\\ \text{d}.& \text{HP}=\{x|-1<x<2,x\in \mathbb{R}\}\\ \text{e}.& \text{HP}=\{x|x<-1\text{atau}x>2,x\in \mathbb{R}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \left|{\displaystyle \frac{x+1}{2}}\right|>\left|{\displaystyle \frac{x-2}{3}}\right|\\ & {\left({\displaystyle \frac{x+1}{2}}\right)}^2>{\left({\displaystyle \frac{x-2}{3}}\right)}^2\\ & \left({\displaystyle \frac{x+1}{2}+\frac{x-2}{3}}\right)\left({\displaystyle \frac{x+1}{2}-{\displaystyle \frac{x-2}{3}}}\right)>0\\ & \left({\displaystyle \frac{3(x+1)+2(x-2)}{6}}\right)\left({\displaystyle \frac{3(x+1)-2(x-2)}{6}}\right)>0\\ & \left({\displaystyle \frac{5x-1}{6}}\right)\left({\displaystyle \frac{x+7}{6}}\right)>0\\ & \text{HP}=\{x|x<-7\text{atau}x>{\displaystyle \frac{1}{5},x\in \mathbb{R}}\}\end{array}\end{array}$.

$\begin{array}{l}\\ 34.& \text{Penyelesaian dari pertidaksamaan}\\ & \left|{\displaystyle \frac{3-2x}{-5}}\right|>5\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-11\text{atau}x>14\\ \text{b}.& x<-14\text{atau}x>11\\ \text{c}.& 11<x<14\\ \text{d}.& -14<x<-11\\ \text{e}.& x>14\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \left|{\displaystyle \frac{3-2x}{-5}}\right|>5\\ & {\displaystyle \frac{3-2x}{-5}<-5\text{atau}{\displaystyle \frac{3-2x}{-5}>5}}\\ & {\displaystyle \frac{2x-3}{5}>5\text{atau}{\displaystyle \frac{2x-3}{5}<-5}}\\ & 2x-3>25\text{atau}2x-3<-25\\ & 2x>25+3\text{atau}2x<-25+3\\ & x>14\text{atau}x<-11,\\ & \text{dapat juga dituliskan}\\ & x<-11\text{atau}x>14\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |2-2|x+1||>4\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-4\text{atau}x>2\\ \text{b}.& x<-3\text{atau}x>1\\ \text{c}.& x<-2\text{atau}x>0\\ \text{d}.& x<-1\text{atau}x>3\\ \text{e}.& x<0\text{atau}x>4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& |2-2|x+1||>4\\ & 2-2|x+1|<-4\text{atau}2-2|x+1|>4\\ & -2|x+1|<-6\text{atau}-2|x+1|>2\\ & |x+1|>3\text{atau}|x+1|<-1\\ & \{\begin{array}{c}(x+1)<-3\\ (x+1)>3\end{array}\text{atau}\{\begin{array}{c}|x+1|<-1\\ \mathbf{\text{tak mungkin}}\end{array}\\ & \text{Selanjutnya}\text{akan didapatkan}\\ & x<-4\text{atau}x>2\end{array}\end{array}$

Latihan Soal 3 Persiapan PAS Gasal Matematika Wajib Kelas X

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

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

$\begin{array}{l}\\ 18.& \text{Penyelesaian yang memenuhi untuk}\\ & |3x+1|=2x+9\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{-2\}& & \text{d}.& \left\{\right\}\\ \text{b}.& \left\{8\right\}& & \text{e}.& \text{setiap bilangan real}\\ \text{c}.& \{-2,8\}& \end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& |3x+1|=2x+9\\ & (3x+1)=\pm (2x+9)\\ & (3x+1)=\{\begin{array}{l}+(2x+9)\\ \\ -(2x+9)\end{array}\end{array}\\ & \begin{array}{rl}\text{Syarat}& :\\ (3x+1)& \{\begin{array}{l}3x+1\ge 0\iff x\ge -{\displaystyle \frac{1}{3}}\\ \\ 3x+1<0\iff x<-{\displaystyle \frac{1}{3}}\end{array}\\ & \end{array}\\ & \begin{array}{|cc|}\hline x\ge -{\displaystyle \frac{1}{3}}& x<-{\displaystyle \frac{1}{3}}\\ \text{Proses}& \text{Proses}\\ \begin{array}{rl}(3x+1)& =+(2x+9)\\ 3x-2x& =9-1\\ x& =8\\ & \end{array}& \begin{array}{rl}(3x+1)& =-(2x+9)\\ 3x+2x& =-9-1\\ 5x& =-10\\ x& =-2\end{array}\\ \mathbf{\text{memenuhi}}& \mathbf{\text{memenuhi}}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Jumlah akar-akar dari}{x}^2+\left|x\right|-6=0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& 4\\ \\ & & & (\mathbf{\text{Entrance Examination}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{x}^2+\left|x\right|-6& =0\\ (\left|x\right|+3)(\left|x\right|-2)& =0\\ \left|x\right|+3=0\text{atau}\left|x\right|-2& =0\\ \left|x\right|=-3(\mathbf{\text{tm}})\text{atau}\left|x\right|& =2(\mathbf{\text{mm}})\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}x& =\pm 2\{\begin{array}{ll}{x}_1& =2\\ x_2& =-2\end{array}\\ & \text{untuk jumlah}\text{dari akar-akarnya adalah}:\\ & x_1+x_2=2+(-2)\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Penyelesaian pertidaksamaan}{x}^2+\left|x\right|-6\le 0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2\le x<0\\ \text{b}.& 0\le x\le 2\\ \text{c}.& -2\le x\le 2\\ \text{d}.& -3\le x\le 2\\ \text{e}.& -2\le x\le 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ \\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}& \text{Proses penyelesaian dipecah jadi 2 bagian}\\ & \text{yaitu}:\{\begin{array}{ll}x& \ge 0\\ x& <0\end{array}\\ & \text{Diketahui pertidaksamaan}:x^2+\left|x\right|-6\le 0\\ & \begin{array}{|cc|}\hline (1)& (2)\\ x\ge 0& x<0\\ \text{maka}\left|x\right|=x& \text{maka}\left|x\right|=-x\\ \begin{array}{rl}& x^2+(x)-6\le 0\\ & x^2+x-6\le 0\\ & (x+3)(x-2)\le 0\\ & \text{Selesaian}:\\ & -3\le x\le 2\\ & \text{karena}x\ge 0,\\ & \text{maka penyelesaian}\\ & \text{menjadi}\\ & 0\le x\le 2\end{array}& \begin{array}{rl}& x^2+(-x)-6\le 0\\ & x^2-x-6\le 0\\ & (x+2)(x-3)\le 0\\ & \text{Selesaian}:\\ & -2\le x\le 3\\ & \text{karena}x<0,\\ & \text{maka penyelesaian}\\ & \text{menjadi}\\ & -2\le x<0\end{array}\\ \hline\end{array}\\ & \text{Gabungan dari penyelesaian (1) dan (2)}\\ & \text{adalah}:-2\le x\le 2\end{array}\\ \\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& \text{Diketahui pertidaksamaan}:x^2+\left|x\right|-6\le 0\\ & \text{dan perlu diingat pula bahwa}:\left|x\right|\ge 0\\ & \text{diubah menjadi}:{\left|x\right|}^2+\left|x\right|-6\le 0\\ & \iff (\left|x\right|+3)(\left|x\right|-2)\le 0\\ & \iff -3\le \left|x\right|\le 2\\ & \text{karena}\left|x\right|\ge 0,\text{maka}\\ & 0\le \left|x\right|\le 2\Rightarrow \left|x\right|\le 2\\ & \text{Sehingga penyelesaian menjadi}\\ & -2\le x\le 2\end{array}\end{array}$.

$\begin{array}{l}\\ 21.& \text{Seluruh bilangan bilangan real}x\\ & \text{yang jaraknya terhadap 3 kurang dari 1 }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3<x<4& \\ \text{b}.& 2<x<3\\ \text{c}.& 2<x<4\\ \text{d}.& 3<x<5\\ \text{e}.& 1<x<3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Seluruh }\text{bilangan bilangan real}x\\ & \text{yang jaraknya terhadap 3 kurang dari 1, }\\ & \text{maksudnya adalah}:\\ & |x-3|<1\\ & \iff -1<x-3<1\\ & \iff -1+\mathbf{\text{(3)}}<x-3+\mathbf{\text{(3)}}<1+\mathbf{\text{(3)}}\\ & \iff 2<x<4\end{array}\end{array}$.

$\begin{array}{l}\\ 22.& \text{Pernyataan berikut yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{Semua nilai}m\text{yang memenuhi}\\ & -1<m<2\text{akan memenuhi juga}\left|m\right|<2\\ \text{b}.& \text{Semua nilai}m\text{yang memenuhi}\\ & -1<m<2\text{akan memenuhi juga}\left|m\right|<1\\ \text{c}.& \text{Semua nilai}m\text{yang memenuhi}\\ & -3<m<-2\text{akan memenuhi juga}\left|m\right|<2\\ \text{d}.& \text{Semua nilai}m\text{yang memenuhi}\\ & 2<m<3\text{akan memenuhi juga}\left|m\right|<2\\ \text{e}.& \text{pilihan jawaban baik a, b, c,}\\ & \text{maupun d tidak ada yang benar}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Perhat}& \text{ikanlah opsi}\mathbf{\text{a}},\\ \left|m\right|& <2\\ -2<m& <2\\ \text{sehing}& \text{ga untuk nilai}m\in \mathbb{R}\text{pada rentang}\\ -1<m& <2\text{akan memenuhi semua}\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Nilai}x\text{real yang memenuhi untuk}\\ & |2x-9|<3\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -3\le x\le 6& \\ \text{b}.& -3<x<6\\ \text{c}.& 3<x<6\\ \text{d}.& 3\le x\le 6\\ \text{e}.& -3<x<-6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& |2x-9|<3\\ & \iff -3<2x-9<3\\ & -3+(9)<2x-9+(9)<3+(9)\\ & \text{masing-masing ditambah 9}\\ & \text{dan akan menjadi bentuk}\\ & 6<2x<12\\ & 6.\left({\displaystyle \frac{1}{2}}\right)<2x.\left({\displaystyle \frac{1}{2}}\right)<12.\left({\displaystyle \frac{1}{2}}\right)\\ & \text{masing-masing dikali}{\displaystyle \frac{1}{2}}\\ & \text{dan akan berubah menjadi bentuk}\\ & 3<x<6\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Nilai}x\text{real yang memenuhi untuk}\\ & |3x+5|\ge 19\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x\le -{\displaystyle \frac{14}{3}\text{atau}x\ge 8}& \\ \\ \text{b}.& x<-8\text{atau}x>{\displaystyle \frac{14}{3}}\\ \\ \text{c}.& x\le -8\text{atau}x\ge {\displaystyle \frac{14}{3}}\\ \\ \text{d}.& x<-{\displaystyle \frac{14}{3}\text{atau}x>8}\\ \\ \text{e}.& x\le 8\text{atau}x\ge -{\displaystyle \frac{14}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& |3x+5|\ge 19\\ \\ & (\ast )-19\ge 3x+5\text{atau}& (\ast \ast )3x+5\ge 19\\ & -19-5\ge 3x\text{atau}& 3x\ge 19-5\\ & -{\displaystyle \frac{24}{3}\ge x\text{atau}}& x\ge {\displaystyle \frac{14}{3}}\\ & -8\ge x\text{atau}& x\ge {\displaystyle \frac{14}{3}}\\ & x\le -8\text{atau}& x\ge {\displaystyle \frac{14}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Nilai}x\text{real yang memenuhi}\\ & 25-|10x+5|\ge |40x-20|\text{adalah}....\\ \\ & (\mathbf{\text{NUS Entrance Examination A level}})\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}:\\ & \begin{array}{rl}& 25-|10x+5|\ge |40x-20|\\ & 25-5|2x+1|\ge 20|2x-1|\\ & 5-|2x+1|\ge 4|2x-1|\\ & \text{ilustrasinya}\begin{array}{lllllll}& & & & & & \\ & & -\frac{1}{2}& & & \frac{1}{2}& & \end{array}\\ & \text{dan berikut}\text{pembagian wilayahnya}\\ & \begin{array}{|ccc|}\hline -\mathrm{\infty }<x<-{\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\le x<\frac{1}{2}}& {\displaystyle \frac{1}{2}\le x<\mathrm{\infty }}\\ \{\begin{array}{ll}|2x+1|& =-(2x+1)\\ |2x-1|& =-(2x-1)\end{array}& \{\begin{array}{ll}|2x+1|& =+(2x+1)\\ |2x-1|& =-(2x-1)\end{array}& \{\begin{array}{ll}|2x+1|& =+(2x+1)\\ |2x-1|& =+(2x-1)\end{array}\\ \hline\end{array}\\ & \text{Selanjutnya adalah}\end{array}\end{array}$

$.\begin{array}{|ccc|}\hline \begin{array}{rl}-\mathrm{\infty }<x<& -\frac{1}{2},\\ 25-|10x+5|& \ge |40x-20|\\ 25-5|2x+1|& \ge 20|2x-1|\\ 5-(-(2x+1))& \ge 4(-(2x-1))\\ 5+2x+1& \ge -8x+4\\ 10x& \ge -2\\ x& \ge -\frac{2}{10}(\mathbf{\text{tm}})\\ & \end{array}& \begin{array}{rl}-\frac{1}{2}\le x<& \frac{1}{2},\\ 25-5|2x+1|& \ge 20|2x-1|\\ 5-(2x+1)& \ge 4(-(2x-1))\\ 5-2x-1& \ge -8x+4\\ 6x& \ge 0\\ x& \ge 0(\mathbf{\text{mm}})\\ \text{yang memenuhi}& =\{0\le x<\frac{1}{2}\}\end{array}& \begin{array}{rl}\frac{1}{2}\le x<& \mathrm{\infty },\\ 25-5|2x+1|& \ge 20|2x-1|\\ 5-(2x+1))& \ge 4(2x-1)\\ 5-2x-1& \ge 8x-4\\ -10x& \ge -8\\ x& \le \frac{8}{10}(\mathbf{\text{mm}})\\ \text{yang memenuhi}& =\{\frac{1}{2}\le x\le \frac{4}{5}\}\end{array}\\ \hline\end{array}$

$.\begin{array}{rl}& \\ & \text{Sehingga yang memenuhi}\text{adalah}:\\ & =\{0\le x\le {\displaystyle \frac{4}{5}}\}\end{array}$.

sumber soal di sini dan di sini

Latihan Soal 2 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 11.& \text{Perhatikanlah ilustrasi grafik di bawah ini}\\ & \begin{array}{}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 12.& \text{Gambarlah garfik untuk persamaan}\\ & \left|x\right|+\left|y\right|=4\\ \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Dari soal diketahui}\\ & \left|x\right|+\left|y\right|=4\\ & \text{maka untuk}\end{array}\\ & \begin{array}{|cccc|}\hline x>0,y>0& & & x>0,y<0\\ x+y=4& & & x+(-y)=4\\ & & & \\ & & & \\ x<0,y>0& & & x<0,y<0\\ (-x)+y=4& & & (-x)+(-y)=4\\ \hline\end{array}\end{array}$.

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

$\begin{array}{l}\\ 13.& \text{Tentukanlah nilai}x\text{dan}y\\ & \text{yang memenuhi}x+\left|x\right|+y=5\\ & \text{dan}x+\left|y\right|-y=10\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & \begin{array}{rl}& \left|x\right|+x+y=5\\ & \text{dan}\\ & x+\left|y\right|-y=10\\ & \end{array}\end{array}\end{array}$.

$.\begin{array}{r}\begin{array}{rl}& \text{untuk}:x>0,y>0\mathbf{\text{(kuadran I)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (x)+x+y=5\\ & \iff 2x+y=5(\text{ada})\\ x+\left|y\right|-y=10& \iff x+(y)-y=10\\ & \iff x=10(\text{ada})\end{array}\\ & \text{untuk}:x<0,y>0\mathbf{\text{(kuadran II)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (-x)+x+y=5\\ & \iff y=5(\text{ada})\\ x+\left|y\right|-y=10& \iff x+(y)-y=10\\ & \iff x=10\\ & \mathbf{\text{(tidak memenuhi)}}\end{array}\end{array}\end{array}$.

$.\begin{array}{r}\begin{array}{rl}& \text{untuk}:x<0,y<0\mathbf{\text{(kuadran III)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (-x)+x+y=5\\ & \iff y=5\\ & \mathbf{\text{(tidak memenuhi)}}\\ x+\left|y\right|-y=10& \iff x+(-y)-y=10\\ & \iff x-2y=10(\text{ada})\end{array}\\ & \text{untuk}:x>0,y<0\mathbf{\text{(kuadran IV)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (x)+x+y=5\\ & \iff 2x+y=5(\text{ada})\\ x+\left|y\right|-y=10& \iff x+(-y)-y=10\\ & \iff x-2y=10(\text{ada})\end{array}\end{array}\end{array}$.

$.\begin{array}{l}\\ & \text{Sebagai ilustrasinya, berikut grafiknya}\\ & \begin{array}{}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Gambarlah grafik fungsi mutlak berikut}\\ & \text{a}.y=|x-2|\\ & \text{b}.y=-|x-2|\\ & \text{c}.y=2+|x-2|\\ & \text{d}.y=2-|x-2|\\ & \text{e}.y=|2+|x-2||\\ & \text{f}.y=|2-|x-2||\end{array}$.

$.\begin{array}{l}\\ & \mathbf{\text{Jawab}}:\\ & \text{Berikut ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{a}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan berikut ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{b}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan berikut pula ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{c}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan berikut pula ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{d}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan sebagai ilustrasinya}\\ & \text{lihat soal no}\mathbf{\text{c}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan terakhir berikut ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{f}}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Penyelesaian yang memenuhi untuk}\\ & |3x-(4x-7)|=6\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{-13,-1\}& & & \text{d}.& \{-13,1\}\\ \text{b}.& \{-1,13\}& \text{c}.& \{1,13\}& \text{e}.& \left\{13\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}|3x-(4x-7)|& =6\\ |3x-4x+7|& =6\\ |-x+7|& =6\\ (-x+7)& =\pm 6\\ -x+7& =\{\begin{array}{l}6\iff -x=6-7\iff x=1\\ \\ -6\iff -x=-6-7\iff x=13\end{array}\end{array}\end{array}$.

Latihan Soal 1 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 1.& \text{Nilai untuk}-|-2021|=....\\ & \begin{array}{l}\\ \text{a}.& -2021& & & \text{d}.& -{2021}^{-1}\\ \text{b}.& 2021& \text{c}.& {2021}^{-1}& \text{e}.& {2021}^2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}-|-2021|& =-\left(2021\right)=-2021\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Nilai untuk}|-4|-|-6^2\times 2|=....\\ & \begin{array}{l}\\ \text{a}.& -68& & & \text{d}.& 68\\ \text{b}.& -40& \text{c}.& 40& \text{e}.& 76\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}|-4|-|-6^2\times 2|& =\left(4\right)-|-36\times 2|\\ & =4-|-72|\\ & =4-\left(72\right)\\ & =-68\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Nilai untuk}(-2022)-|-(-2021)|-{|-3^2|}^2=....\\ & \begin{array}{l}\\ \text{a}.& -3960& & & \text{d}.& -4068\\ \text{b}.& -4038& \text{c}.& -4050& \text{e}.& -4124\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& (-2022)-|-(-2021)|-{|-3^2|}^2\\ & =-2022-2021-{|-9|}^2\\ & =-4043-9^2\\ & =-4043-81\\ & =-4124\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai untuk}{|-4|}^{|-2|}-{|-2|}^{|-4|}=....\\ & \begin{array}{l}\\ \text{a}.& -32& & & \text{d}.& 16\\ \text{b}.& -16& \text{c}.& 0& \text{e}.& 32\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{|-4|}^{|-2|}-{|-2|}^{|-4|}& =4^2-2^4\\ & =16-16\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai untuk}\\ & |1-2^2|+|2^2-3^2|+|3^2-4^2|+\cdots +|{2020}^2-{2021}^2|....\\ & \begin{array}{l}\\ \text{a}.& -{2021}^2& & & \text{d}.& {2021}^2-1\\ \text{b}.& 1-{2021}^2& \text{c}.& -2021& \text{e}.& {2021}^2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& |1-2^2|+|2^2-3^2|+|3^2-4^2|+\cdots +|{2020}^2-{2021}^2|\\ & =(2^2-1)+(3^2-2^2)+(4^2-3^2)+\cdots +({2021}^2-{2020}^2)\\ & =-1+{2021}^2\\ & ={2021}^2-1\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Nilai untuk}\\ & \left|{\displaystyle \frac{1}{2}-1}\right|\times \left|{\displaystyle \frac{1}{3}-1}\right|\times \left|{\displaystyle \frac{1}{4}-1}\right|\times \left|{\displaystyle \frac{1}{5}-1}\right|\times \cdots \times \left|{\displaystyle \frac{1}{2021}-1}\right|=....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2021}}& & & \text{d}.& {\displaystyle \frac{2020}{2021}}\\ \\ \text{b}.& -{\displaystyle \frac{2020}{2021}}& \text{c}.& -2021{\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle \frac{1}{2021}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \left|{\displaystyle \frac{1}{2}-1}\right|\times |\frac{1}{3}-1|\times |\frac{1}{4}-1|\times |\frac{1}{5}-1|\times \cdots \times |\frac{1}{2021}-1|\\ & =|-{\displaystyle \frac{1}{2}}|\times |-\frac{2}{3}|\times |-\frac{3}{4}|\times |-\frac{4}{5}|\times \cdots \times |-\frac{2020}{2021}|\\ & ={\displaystyle \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \cdots \times \frac{2019}{2020}\times \frac{2020}{2021}}\\ & ={\displaystyle \frac{1}{2021}}\end{array}\end{array}$.

$\begin{array}{ll}7.& \text{Bentuk sederhana dari}x-5y\text{dan}\\ & 5y-x\text{adalah}....\\ & \begin{array}{llll}\text{a}.& |5y-x|& \text{d}.& |y-x|\\ \text{b}.& |y-5x|& \text{e}.& |-5y-x|\\ \text{c}.& |x-5y|\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & |x-5y|=m=\{\begin{array}{ll}\bullet & =x-5y\\ \bullet & =-(x-5y)\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Nilai}m\text{yang memenuhi}\left|4m\right|=16\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 4\\ \text{c}.& \pm 2\\ \text{d}.& \pm 4\\ \text{e}.& \pm 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left|4m\right|& =16\\ (4m)& =\pm 16\\ m& =\pm {\displaystyle \frac{16}{4}}\\ & =\pm 4\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Nilai}x\text{yang memenuhi untuk}|2x+5|=9\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2& & \text{d}.& -7\text{dan}2\\ \text{b}.& 2\text{dan}7& & \text{e}.& -2\text{dan}7\\ \text{c}.& -7\text{dan}-2& \end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}|2x+5|& =9\\ (2x+5)& =\pm 9\\ 2x& =\pm 9-5\\ x& ={\displaystyle \frac{\pm 9-5}{2}}\\ x& =\{\begin{array}{l}={\displaystyle \frac{+9-5}{2}=\frac{4}{2}=2}\\ \\ ={\displaystyle \frac{-9-5}{2}=\frac{-14}{2}=-7}\end{array}\end{array}\end{array}$.

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

Problem Solving Bentuk Bilangan Riil

Seri Pemecahan Masalah

Jika pada bahasan sebelumnya kita bahas bilangan tidak nyata atau bilangan imajiner pada akar persamaan kuadrat, sekarang kita ketengahkan bahasan sebaliknya, yaitu akar nyta atau riil dari suatu persamaan kuadrat. 

Berikut permasalahannya

(sumber soal dari blog saya sendiri di wordpress)

$\begin{array}{rl}& \text{Akar riil terbesar untuk persamaan}\\ & {\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4}\\ & \text{adalah}p+\sqrt{q+\sqrt{r}}\text{dengan}p,q,\text{dan}r\text{adalah}\\ & \text{bilangan asli}.\text{Tentukanlah nilai}p+q+r\\ \\ & \mathbf{\text{Solusi}}:\end{array}$.

$\begin{array}{rl}& {\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4}\\ & \frac{3}{x-3}+1+\frac{5}{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1=x^2-11x\\ & \frac{3+(x-3)}{x-3}+\frac{5+(x-5)}{x-5}+\frac{17+(x-17)}{x-17}+\frac{19+(x-19)}{x-19}=x^2-11x\\ & \frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}=x^2-11x\\ & \frac{x(x-19)+x(x-3)}{(x-3)(x-19)}+\frac{x(x-17)+x(x-5)}{(x-5)(x-17)}=x^2-11x\\ & \frac{2x^2-22x}{x^2-22x+57}+\frac{2x^2-22}{x^2-22x+85}=x^2-11x\\ & (x^2-11x)(\frac{2}{x^2-22x+57}+\frac{2}{x^2-22x+85})\\ & =x^2-11x,\text{misal}t=x^2-22x\\ & (\frac{2}{t+57}+\frac{2}{t+85})=\frac{x^2-11x}{x^2-11x}=1\\ & 2(t+85)+2(t+57)=(t+57)(t+85)\\ & 2t+170+2t+114=t^2+142t+4845\\ & 0=t^2+138t+4731\\ & t^2+138t+4731=0\{\begin{array}{c}a=1\\ b=138\\ c=4731\end{array}\\ & t_{1,2}={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\\ & t_{1,2}={\displaystyle \frac{-138\pm \sqrt{{138}^2-4.1.4731}}{2}}\\ & ={\displaystyle \frac{-138\pm \sqrt{19044-18924}}{2}}\\ & ={\displaystyle \frac{-138\pm \sqrt{120}}{2}}\\ & ={\displaystyle \frac{-138\pm 2\sqrt{30}}{2}}\\ & =-69\pm \sqrt{30}\end{array}$.

$\begin{array}{rl}& \text{Selanjutnya}\\ & t_{1,2}=-69\pm \sqrt{30}\\ & x^2-22x=-69\pm \sqrt{30}\\ & x^2-22x+69\pm \sqrt{30}=0\\ & x^2-22x+69+\sqrt{30}=0\\ & \text{atau}{x}^2-22x+69-\sqrt{30}\\ & \\ & \text{dengan cara yang}\text{semisal diatas}\\ & \\ & x_{1,2}={\displaystyle \frac{22\pm \sqrt{{22}^2-4(69+\sqrt{30})}}{2}}\\ & \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm \sqrt{{22}^2-4(69-\sqrt{30})}}{2}}\\ & x_{1,2}={\displaystyle \frac{22\pm \sqrt{484-276-4\sqrt{30}}}{2}}\\ & \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm \sqrt{484-276+4\sqrt{30}}}{2}}\\ & x_{1,2}={\displaystyle \frac{22\pm \sqrt{208-4\sqrt{30}}}{2}}\\ & \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm \sqrt{208+4\sqrt{30}}}{2}}\\ & x_{1,2}={\displaystyle \frac{22\pm 2\sqrt{52-\sqrt{30}}}{2}}\\ & \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm 2\sqrt{52+\sqrt{30}}}{2}}\\ & x_{1,2}=11\pm \sqrt{52-\sqrt{30}}\\ & \text{atau}{x}_{3,4}=11\pm \sqrt{52+\sqrt{30}}\\ & \\ & \text{Maka},\\ & \{\begin{array}{c}{x}_1=11+\sqrt{52-\sqrt{30}}\\ \\ x_2=11-\sqrt{52-\sqrt{30}}\end{array}\\ & \text{atau}\{\begin{array}{c}{x}_3=11+\sqrt{52+\sqrt{30}}\\ \\ x_4=11-\sqrt{52+\sqrt{30}}\end{array}\end{array}$.

$\begin{array}{rl}\text{Selanjutnya nilai}& \text{yang paling pas sesuai soal adalah}:\\ & x_3=11+\sqrt{52+\sqrt{30}}=p+\sqrt{q+\sqrt{r}}\\ \text{Sehingga nilai}& p+q+r=11+52+30=93\end{array}$.

Problem Solving Bentuk Bilangan Imajiner (Bilangan Tidak Nyata)

Seri Pemecahan Masalah

Suatu ketika saya sharing-sharing mengenai soal bentuk perpangkatan dari salah seorang teman yang kebetulan memang soalnya membuat penasayaran untuk ditemukan jawabannya.

Berikut soalnya

Saat saya melihat soalnya dengan pangkat berupa angka yang seolah berpola tapi agak susah dicari hungan antara keduanya. Yang satu bilangan utuh yang satu lagi bentuk pecahan (bilangan pada soal, bukan pada yang diketahui). Tapi ada sedikit petunjuk yang mensiratkan soal di atas akan segera dapat dipecahkankan, yaitu posisi yang diketahui  $x+{\displaystyle \frac{1}{x}=-1}$ adalah salah satu bentuk persamaan kuadrat dengan akar kemungkinan rasional atau imajiner/khayal/tidak nyata dan pangkat pada soal yang semuanya menunjukkan kelipatan 3, yaitu 1234567891011 dan yang satunya posisi penyebut dengan pangkat 1110987654321 dengan basis/bilangan pokok perpangkatannya sama dengan yang diketahui dari soal yaitu  $a$.

Sebelumnya saya pernah menyinggung mengenai istilah definit positif dan definit negatif (silahkan klik di sini) yang kurang lebih istilah tersebut sangat berkaitan dengan akar persamaan kuadrat yang berbentuk imajiner.

Ok, kita kembali ke arah penyelesaian soal di atas, yaitu:

$\begin{array}{rl}& a+{\displaystyle \frac{1}{a}=-1\iff a^2+1=-a}\\ & \iff a^2+a+1=0\\ & \iff a_{1,2}={\displaystyle \frac{-1\pm \sqrt{-3}}{2}={\displaystyle \frac{-1\pm \sqrt{3.(-1)}}{2}}}\\ & ={\displaystyle \frac{-1\pm \sqrt{3}\sqrt{-1}}{2}=\frac{-1\pm \sqrt{3}i}{2}}\\ & \text{dengan}i=\sqrt{-1}\end{array}$.

$\begin{array}{rl}& \text{Misalkan kita pilih}a={\displaystyle \frac{-1+\sqrt{3}i}{2}}\\ & \text{maka nilai dari}\\ & {\displaystyle \frac{1}{a}={\displaystyle \frac{1}{{\displaystyle \frac{-1+\sqrt{3}i}{2}}}={\displaystyle \frac{2}{-1+\sqrt{3}i}={\displaystyle \frac{2}{\sqrt{3}i-1}}}}}\\ & ={\displaystyle \frac{2}{\sqrt{3}i-1}\times {\displaystyle \frac{\sqrt{3}i+1}{\sqrt{3}i+1}={\displaystyle \frac{2(\sqrt{3}i+1)}{-3-1}}}}\\ & =-{\displaystyle \frac{2(\sqrt{3}i+1)}{-4}={\displaystyle \frac{\sqrt{3}i+1}{-2}\text{atau}}}\\ & {\displaystyle \frac{1}{a}={\displaystyle \frac{-1-\sqrt{3}i}{2}}}\end{array}$.

Penjabaran bentuk pangkat dari salah satu akar ternyata membentuk pola yang unik sebagaimana bentuk berikut:

$\begin{array}{rl}& \{\begin{array}{ll}a& ={\displaystyle \frac{-1+\sqrt{3}i}{2}}\\ {\displaystyle \frac{1}{a}}& ={\displaystyle \frac{-1-\sqrt{3}i}{2}}\end{array},\{\begin{array}{ll}{a}^2& ={\displaystyle \frac{-1-\sqrt{3}i}{2}}\\ {\displaystyle \frac{1}{a^2}}& ={\displaystyle \frac{-1+\sqrt{3}i}{2}}\end{array}\\ & \{\begin{array}{ll}{a}^3& =1\\ {\displaystyle \frac{1}{a^3}}& =1\end{array}\\ & \{\begin{array}{ll}{a}^4& ={\displaystyle \frac{-1+\sqrt{3}i}{2}}\\ {\displaystyle \frac{1}{a^4}}& ={\displaystyle \frac{-1-\sqrt{3}i}{2}}\end{array}\{\begin{array}{ll}{a}^5& ={\displaystyle \frac{-1-\sqrt{3}i}{2}}\\ {\displaystyle \frac{1}{a^5}}& ={\displaystyle \frac{-1+\sqrt{3}i}{2}}\end{array}\\ & \{\begin{array}{ll}{a}^6& =1\\ {\displaystyle \frac{1}{a^6}}& =1\end{array}\\ & ⋮\\ & \cdots \cdots \{\begin{array}{ll}{a}^9& =1\\ {\displaystyle \frac{1}{a^9}}& =1\end{array}\\ & \cdots \cdots \{\begin{array}{ll}{a}^{12}& =1\\ {\displaystyle \frac{1}{a^{12}}}& =1\end{array}\\ & \cdots \cdots \{\begin{array}{ll}{a}^{15}& =1\\ {\displaystyle \frac{1}{a^{15}}}& =1\end{array}\\ & \text{dan seterusnya}\\ & \end{array}$.

Jadi, setiap pangkat kelipatan 3 ternyata sama dengan 1, sehingga ini mengakibatkan soal di atas dapat dituliskan lagi dengan

$\begin{array}{rl}& \text{Perhatikan lagi bentuk soal}\\ & a^{1234567891011}+{\displaystyle \frac{1}{a^{11100987654321}}}\\ & =a^{3m}+{\displaystyle \frac{1}{a^{3n}}=1+{\displaystyle \frac{1}{1}=1+1=2}}\end{array}$.

Contoh Soal 13 Statistika

$\begin{array}{ll}56.& \text{Simpangan baku dari data berikut}:\\ & 6,7,4,5,3\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& {\displaystyle \frac{1}{2}}& & & \text{d}.& \sqrt{2}\\ \text{b}.& {\displaystyle \frac{1}{2}\sqrt{2}}& \text{c}.& {\displaystyle \frac{1}{2}\sqrt{3}}& \text{e}.& \sqrt{3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 6,7,4,5,3\\ & \text{Simpangan bakuya adalah}:\\ & S=\sqrt{S^2}=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{6+7+4+5+3}{5}={\displaystyle \frac{25}{5}=5}}\\ & \text{maka nilai}\\ & S=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & =\sqrt{{\displaystyle \frac{1}{5}((6-5{)}^2+(7-5{)}^2+(4-5{)}^2+(5-5{)}^2+(3-5{)}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(1^2+2^2+1^2+0+2^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(1+4+1+4)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(10)}}=\sqrt{{\displaystyle \frac{10}{5}}}=\sqrt{2}\end{array}\end{array}$.

$\begin{array}{ll}57.& \mathbf{\text{UN 2010}}\\ & \text{Simpangan baku dari data berikut}:\\ & 2,3,4,5,6\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& \sqrt{15}& & & \text{d}.& \sqrt{3}\\ \text{b}.& {\displaystyle \sqrt{10}}& \text{c}.& {\displaystyle \sqrt{5}}& \text{e}.& \sqrt{2}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 2,3,4,5,6\\ & \text{Simpangan bakuya adalah}:\\ & S=\sqrt{S^2}=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{2+3+4+5+6}{5}={\displaystyle \frac{20}{5}=4}}\\ & \text{maka nilai}\\ & S=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & =\sqrt{{\displaystyle \frac{1}{5}((2-5{)}^2+(3-5{)}^2+(4-5{)}^2+(5-5{)}^2+(6-5{)}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(3^2+2^2+1^2+0+1^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(9+4+1+1)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(15)}}=\sqrt{{\displaystyle \frac{15}{5}}}=\sqrt{3}\end{array}\end{array}$.

$\begin{array}{ll}58.& \text{Simpangan baku dari data berikut}:\\ & 7,9,11,13,15\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& 2,4& & & \text{d}.& 2,8\\ \text{b}.& {\displaystyle 2,5}& \text{c}.& {\displaystyle 2,7}& \text{e}.& 2,9\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 7,9,11,13,15\\ & \text{Simpangan bakuya adalah}:\\ & S=\sqrt{S^2}=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{7+9+11+13+15}{5}={\displaystyle \frac{55}{5}=11}}\\ & \text{maka nilai}\\ & S=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & =\sqrt{{\displaystyle \frac{1}{5}((7-11{)}^2+(9-11{)}^2+(11-11{)}^2+(13-11{)}^2+(15-11{)}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(4^2+2^2+0+2^2+4^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(16+4+4+16)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(40)}}=\sqrt{{\displaystyle \frac{40}{5}}}=\sqrt{8}=2,82..\end{array}\end{array}$.

 $\begin{array}{ll}59.& \text{Simpangan baku dari data berikut}:\\ & 2,4,4,5,6,6,7,8,9,9\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& 4\sqrt{3}& & & \text{d}.& {\displaystyle \frac{2}{5}\sqrt{30}}\\ \text{b}.& 2{\displaystyle \frac{2}{5}}& \text{c}.& {\displaystyle \sqrt{5}}& \text{e}.& 2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 2,4,4,5,6,6,7,8,9,9\\ & \text{Simpangan bakunya adalah}:\\ & S=\sqrt{S^2}=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{2+4+4+5+6+6+7+8+9+9}{10}={\displaystyle \frac{60}{10}=6}}\\ & \text{maka nilai}\\ & S=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & =\sqrt{{\displaystyle \frac{1}{10}((2-6{)}^2+2(4-6{)}^2+(5-6{)}^2+2(6-6{)}^2+(7-6{)}^2+(8-6{)}^2+2(9-6{)}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{10}(4^2+{2.2}^2+1^2+0+1^2+2^2+{2.3}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{10}(16+8+1+1+4+18)}}\\ & =\sqrt{{\displaystyle \frac{1}{10}(48)}}=\sqrt{{\displaystyle \frac{48}{10}}}=\sqrt{{\displaystyle \frac{120}{25}}}={\displaystyle \frac{2}{5}\sqrt{30}}\end{array}\end{array}$.

Contoh Soal 12 Statistika

$\begin{array}{ll}51.& \text{Simpangan kuartil dari data}\\ & 56a378\text{adalah}1{\displaystyle \frac{1}{2}.\text{Jika median datanya}}\\ & \text{adalah}5{\displaystyle \frac{1}{2},\text{maka rata-rata data}\text{tersebut adalah}....}\\ & \begin{array}{llllll}\text{a}.& \text{4}& & & \text{d}.& \text{5}{\displaystyle \frac{1}{2}}\\ \text{b}.& \text{4}{\displaystyle \frac{1}{2}}& \text{c}.& \text{5}& \text{e}.& \text{6}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 56a378\Rightarrow n=6\\ & \text{karena mediannya}=M_e=Q_2=5\frac{1}{2}=\frac{11}{2}\\ & \text{ data menjadi}:a35678\text{atau}3a5678\\ & \text{Simpangan kuartilnya adalah}1{\displaystyle \frac{1}{2}={\displaystyle \frac{3}{2}}}\\ & Q_d={\displaystyle \frac{1}{2}(Q_3-Q_1)={\displaystyle \frac{3}{2}\Rightarrow Q_3-Q_1=3}}\\ & \text{maka}\\ & 3=Q_3-Q_1=x_{{.}_{\frac{3}{4}n+\frac{1}{2}}}-x_{{.}_{\frac{1}{4}n+\frac{1}{2}}}=x_{{.}_{\frac{3}{4}6+\frac{1}{2}}}-x_{{.}_{\frac{1}{4}6+\frac{1}{2}}}\\ & =(x_{{.}_5}-x_{{.}_2})=7-x_{{.}_2}=3\Rightarrow x_{{.}_2}=4\\ & \text{Jadi, rata-ratanya adalah}:\\ & \bar{x}=\frac{3+4+5+6+7+8}{6}=5,5\end{array}\end{array}$.

$\begin{array}{ll}52.& \text{Simpangan kuartil dari data}\\ & \text{berikut}\\ & 61,61,50,50,53,53,70,61\\ & 53,70,53,61,50,61,70\\ & \text{adalah}....\\ & \begin{array}{llllll}\text{a}.& \text{10}& & & \text{d}.& \text{6}\\ \text{b}.& \text{9}& \text{c}.& \text{8}& \text{e}.& \text{5}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{-}\\ & \begin{array}{rl}& \text{Data mula-mula}n=15\\ & 61,61,50,50,53,53,70,61\\ & 53,70,53,61,50,61,70\\ & \text{data durutkan}\\ & 50,50,50,53,53,53,53\\ & 61,61,61,61,61,70,70,70\\ & \text{Simpangan kuartil adalah}{Q}_d,\\ & Q_d={\displaystyle \frac{1}{2}(Q_3-Q_1)}\\ & ={\displaystyle \frac{1}{2}(x_{{.}_{\frac{3}{4}(n+1)}}-x_{{.}_{\frac{1}{4}(n+1)}})}\\ & ={\displaystyle \frac{1}{2}(x_{{.}_{\frac{3}{4}(15+1)}}-x_{{.}_{\frac{1}{4}(15+1)}})}\\ & ={\displaystyle \frac{1}{2}(x_{{.}_{12}}-x_{{.}_4})}\\ & ={\displaystyle \frac{1}{2}(61-53)={\displaystyle \frac{1}{2}\times 8=4}}\\ & \text{Jadi},Q_d=4\end{array}\end{array}$.

$\begin{array}{ll}53.& \text{Simpangan rata-rata dari data berikut}:\\ & 642810\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& \text{2}& & & \text{d}.& \text{3},0\\ \text{b}.& \text{2},4& \text{c}.& \text{2},5& \text{e}.& \text{3},5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{b}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 642810\\ & \text{Simpangan rata-ratanya adalah}:\\ & SR={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{6+4+2+8+10}{5}={\displaystyle \frac{30}{5}=6}}\\ & \text{maka nilai}\\ & SR={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & ={\displaystyle \frac{1}{5}(|6-6|+|4-6|+|2-6|+|8-6|+|10-6|)}\\ & ={\displaystyle \frac{1}{5}(\left|0\right|+|-2|+|-4|+\left|2\right|+\left|4\right|)}\\ & ={\displaystyle \frac{1}{5}(0+2+4+2+4)}\\ & ={\displaystyle \frac{1}{5}(12)={\displaystyle \frac{12}{5}=2,4}}\end{array}\end{array}$.

$\begin{array}{ll}54.& \text{Simpangan rata-rata dari data berikut}:\\ & 10,8,7,10,7,5,8,6,10,9\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& \text{1,0}& & & \text{d}.& \text{8,0}\\ \text{b}.& \text{1,4}& \text{c}.& \text{6,0}& \text{e}.& \text{14,0}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{b}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 10,8,7,10,7,5,8,6,10,9\\ & \text{Simpangan rata-ratanya adalah}:\\ & SR={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{10+8+7+10+7+5+8+6+10+9}{10}}\\ & ={\displaystyle \frac{80}{10}=8}\\ & \text{maka nilai}\\ & SR={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & ={\displaystyle \frac{1}{10}(|5-8|+|6-8|+2|7-8|+2|8-8|+|9-8|+3|10-8|)}\\ & ={\displaystyle \frac{1}{10}(|-3|+|-2|+2|-1|+2\left|0\right|+\left|1\right|+3\left|2\right|)}\\ & ={\displaystyle \frac{1}{10}(3+2+2+0+1+6)}\\ & ={\displaystyle \frac{1}{10}(14)={\displaystyle \frac{14}{10}=1,4}}\end{array}\end{array}$.

$\begin{array}{ll}55.& \text{Nilai variansi dari data}\\ & 6,7,7,8,8,8,8,12\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& \text{1}& & & \text{d}.& \text{8}\\ \text{b}.& {\displaystyle \frac{26}{8}}& \text{c}.& {\displaystyle \frac{11}{4}}& \text{e}.& \text{22}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{c}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 6,7,7,8,8,8,8,12\\ & \text{Variannya adalah}:\\ & S^2={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{6+7+7+8+8+8+8+12}{8}={\displaystyle \frac{64}{8}=8}}\\ & \text{maka nilai}\\ & S^2={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}\\ & ={\displaystyle \frac{1}{8}((6-8{)}^2+2(7-6{)}^2+4(8-8{)}^2+(12-8{)}^2)}\\ & ={\displaystyle \frac{1}{8}(2^2+2.1+4.0+4^2)}\\ & ={\displaystyle \frac{1}{8}(4+2+0+16)}\\ & ={\displaystyle \frac{1}{8}(22)={\displaystyle \frac{11}{4}=2,75}}\end{array}\end{array}$.