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

 $\begin{array}{ll}\\ 6.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{2x+6}{x-4}\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-10<x<4\\ \color{red}\textrm{b}.&-10\leq x<4\\ \textrm{c}.&-4<x\leq 10\\ \textrm{d}.&x\leq -10\: \: \textrm{atau}\: \: x\geq 4\\ \textrm{e}.&x<-10\: \: \textrm{atau}\: \: x\geq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\displaystyle \frac{2x+6}{x-4}&\leq 1\\ \displaystyle \frac{2x+6}{x-4}-1&\leq 0\\ \displaystyle \frac{2x+6-(x-4)}{x-4}&\leq 0\\ \displaystyle \frac{x+10}{x-4}&\leq 0\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \displaystyle \frac{x^{2}-x}{x+3}\geq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-3\: \: \textrm{atau}\: \: -1\leq x\leq 3\\ \color{red}\textrm{b}.&-3< x\leq -1\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{c}.&-3\leq x\leq 3\\ \textrm{d}.&-3\leq x\leq -1\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{e}.&-3\leq x\leq -1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x^{2}-x}{x+3}&\geq 1\\ \displaystyle \frac{x^{2}-x}{x+3}&-1\geq 0\\ \displaystyle \frac{x^{2}-x-(x+3)}{x+3}&\geq 0\\ \displaystyle \frac{x^{2}-2x-3}{x+3}&\geq 0\\ \displaystyle \frac{(x-3)(x+1)}{x+3}&\geq 0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x+2+\displaystyle \frac{1}{x+4}>0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-4\: \: \textrm{atau}\: \: x\geq -3\\ \textrm{b}.&x<-4\: \: \textrm{atau}\: \: x>-3\\ \textrm{c}.&-4\leq x\leq -3\\ \color{red}\textrm{d}.&x>-4\\ \textrm{e}.&-4\leq x\leq -3\: \: \textrm{atau}\: \: x>-3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}x+2+\displaystyle \frac{1}{x+4}&>0\\ \displaystyle \frac{(x+2)(x+4)+1}{(x+4)}&>0\\ \displaystyle \frac{x^{2}+6x+8+1}{x+4}&>0\\ \displaystyle \frac{x^{2}+6x+9}{x+4}&>0\\ \displaystyle \frac{(x+3)^{2}}{(x+4)}&>0\\ x&>-4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 9.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x+3<\displaystyle \frac{x^{2}+6x+11}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|x< -3\displaystyle \frac{2}{3}\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \}\\ \textrm{b}.&\left \{ x|0\leq x\leq 11,\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\left \{ x|x<-11\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \}\\ \textrm{d}.&\left \{ x|x<0\: \: \textrm{atau}\: \: x>11,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\left \{ x|x\leq 11\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}x+3<\displaystyle \frac{x^{2}+6x+11}{x}&\\ x+3-\displaystyle \frac{x^{2}+6x+11}{x}&<0\\ \displaystyle \frac{x(x+3)-\left (x^{2}+6x+11 \right )}{x}&<0\\ \displaystyle \frac{x^{2}+3x-x^{2}-6x-11}{x}&<0\\ \displaystyle \frac{-3x-11}{x}&<0\\ \displaystyle \frac{3x+11}{x}&>0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&\textrm{Nilai}\: \: x\: \: \textrm{berikut yang tidak memenuhi}\\ &\displaystyle \frac{x-3}{x^{2}+2x+1}\leq 0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \color{red}\textrm{b}.&-1\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x-3}{x^{2}+2x+1}\leq 0&\\ \displaystyle \frac{(x-3)}{(x+1)^{2}}\leq 0&\\ \textrm{Pembuat nol}&\\ \begin{cases} x & =3 ,\: \textrm{boleh digunakan}\\ x& =-1,\: \textrm{\textrm{tetapi}}\: \: x\neq -1, \end{cases}&\\ \textrm{sehingga}\: \: -1\: \: \textrm{tidak digunakan}&\\ \color{black}\begin{array}{ccc|cccc|ccccc}\\ &&&&&&&&&&\\ &-&-&-&-&-&&+&+&&\\\hline &&-1&&&&3&&&&\\ &&&&&&&&&&\\ \end{array}& \end{aligned} \end{array}$