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

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

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

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

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

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