Latihan Soal 6 Persiapan PAS Gasal Matematika Wajib Kelas X

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

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

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

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

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

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

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

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

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

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