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

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

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

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

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

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