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

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

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

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

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

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