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

$\begin{array}{ll}\\ 1.&\textrm{Diketahui pertidaksamaan}\: \: \displaystyle \frac{x+10}{x-9}\leq 0\\ &\textrm{dan diberikan beberapa nilai berikut}\\ &(\textrm{i})\quad x=-6\: \, \qquad\qquad (\textrm{iii})\quad x=-14\\ &(\textrm{ii})\, \, \, \: x=-10\qquad\quad\quad (\textrm{iv})\quad x=-18\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\textrm{di atas adalah ditunjukkan oleh}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&(\textrm{i})\: \: \textrm{dan} \: \: (\textrm{ii})\\ \textrm{b}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{c}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{d}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iv})\\ \textrm{e}.&(\textrm{iii})\: \: \textrm{dan}\: \: \: (\textrm{iv}) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x+10}{x-9}&\leq 0\\ \textrm{HP}=&\color{red}\left \{ x|-10\leq x< 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\leq x< 6\\ \textrm{b}.&2\leq x< 3\\ \color{red}\textrm{c}.&2< x< 3\: \: \textrm{atau}\: \: x>6\\ \textrm{d}.&x<3\: \: \textrm{atau}\: \: 3<x<6\\ \textrm{e}.&x<2\: \: \textrm{atau}\: \: x>3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\displaystyle \frac{6}{x-3}<\frac{8}{x-2}&\\ \displaystyle \frac{6}{x-3}-\frac{8}{x-2}&<0\\ \displaystyle \frac{6(x-2)-8(x-3)}{(x-3)(x-2)}&<0\\ \displaystyle \frac{6x-8x-12+24}{(x-2)(x-3)}&<0\\ \displaystyle \frac{-2x+12}{(x-2)(x-3)}&<0\\ \displaystyle \frac{2x-12}{(x-2)(x-3)}&>0\\ \displaystyle \frac{2(x-6)}{(x-2)(x-3)}&>0\\ \textrm{HP}=&\color{red}\left \{ x|2<x<3\: \: \textrm{atau}\: \: x>6,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{x^{2}-81}{x^{2}}\geq 0\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x\leq -9\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{b}.&-9\leq x< 0\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{c}.&-9\leq x< 0\: \: \textrm{atau}\: \: 0<x\leq 9\\ \textrm{d}.&-9< x\leq 9\\ \textrm{e}.&x\in \mathbb{R} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x^{2}-81}{x^{2}}&\geq 0\\ \displaystyle \frac{(x+9)(x-9)}{x^{2}}&\geq 0\\ \textrm{HP}=&\color{red}\left \{ x|x\leq -9\: \: \textrm{atau}\: \: x\geq 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 5.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}<0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|-9< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{b}.&\left \{ x|-6< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \}\\ \color{red}\textrm{d}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: \displaystyle \frac{5}{2}<x<5,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\left \{ x|x< -9\: \: \textrm{atau}\: \: -6< x< \displaystyle \frac{5}{2}\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}&<0\\ \displaystyle \frac{(x+6)(x-5)}{(x+9)(2x-5)}&<0\\ \textrm{Cukup jelas}& \end{aligned} \end{array}$