Latihan Soal 6 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 46.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{2}{x+1}\leq \left | x \right |\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{b}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: 0\leq x\leq 1 \right \}\\ \textrm{c}.&\left \{ x|x\geq 1 \right \}\\ \color{red}\textrm{d}.&\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{e}.&\left \{ x|-1< x\leq 1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | x \right |\geq \displaystyle \frac{2}{x+1}\quad\quad\quad \color{black}\textrm{berakibat}\\ &\displaystyle \frac{-2}{x+1}\geq x\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{2}{x+1}\\ &\bullet \qquad \color{red}\textrm{bagian}\: \: 1\\ &x\leq \displaystyle \frac{-2}{x+1}\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ &x+\displaystyle \frac{2}{x+1}\leq 0\\ &\displaystyle \frac{x(x+1)+2}{x+1}\leq 0\\ &\displaystyle \frac{x^{2}+x+2}{x+1}\leq 0\Leftrightarrow \displaystyle \frac{\textrm{Definit positif}}{x+1}\leq 0\\ &\begin{aligned}&\textrm{HP}_{1}=\color{black}\left \{x| x< -1,\: x\in \mathbb{R} \right \}\\ &\bullet \qquad \color{red}\textrm{bagian}\: \: 2\\ &x\geq \displaystyle \frac{2}{x+3}\\ &x-\displaystyle \frac{2}{x+1}\geq 0\\ &\displaystyle \frac{x(x+1)-2}{x+1}\geq 0\\ &\displaystyle \frac{x^{2}+x-2}{x+1}\geq 0\\ &\displaystyle \frac{(x+2)(x-1)}{x+1}\geq 0\\ &\textrm{HP}_{2}=\color{black}\left \{x|-2\leq x< -1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \}\\ &\textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}=\color{red}\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 47.&\textrm{Diketahui pertidaksamaan}\: \: \displaystyle \frac{x+10}{x-9}\leq 0\\ &\textrm{dan diberikan beberapa nilai berikut}\\ &(\textrm{i})\quad x=-6\: \, \qquad\qquad (\textrm{iii})\quad x=-14\\ &(\textrm{ii})\, \, \, \: x=-10\qquad\quad\quad (\textrm{iv})\quad x=-18\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\textrm{di atas adalah ditunjukkan oleh}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&(\textrm{i})\: \: \textrm{dan} \: \: (\textrm{ii})\\ \textrm{b}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{c}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{d}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iv})\\ \textrm{e}.&(\textrm{iii})\: \: \textrm{dan}\: \: \: (\textrm{iv}) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x+10}{x-9}&\leq 0\\ \textrm{HP}=&\color{red}\left \{ x|-10\leq x< 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 48.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\leq x< 6\\ \textrm{b}.&2\leq x< 3\\ \color{red}\textrm{c}.&2< x< 3\: \: \textrm{atau}\: \: x>6\\ \textrm{d}.&x<3\: \: \textrm{atau}\: \: 3<x<6\\ \textrm{e}.&x<2\: \: \textrm{atau}\: \: x>3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\\ &\Leftrightarrow \displaystyle \frac{6}{x-3}-\frac{8}{x-2}<0\\ &\Leftrightarrow \displaystyle \frac{6(x-2)-8(x-3)}{(x-3)(x-2)}<0\\ &\Leftrightarrow \displaystyle \frac{6x-8x-12+24}{(x-2)(x-3)}<0\\ &\Leftrightarrow \displaystyle \frac{-2x+12}{(x-2)(x-3)}<0\\ &\Leftrightarrow \displaystyle \frac{2x-12}{(x-2)(x-3)}>0\\ &\Leftrightarrow \displaystyle \frac{2(x-6)}{(x-2)(x-3)}>0\\ &\textrm{HP}=\color{red}\left \{ x|2<x<3\: \: \textrm{atau}\: \: x>6,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 49.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{x^{2}-81}{x^{2}}\geq 0\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x\leq -9\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{b}.&-9\leq x< 0\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{c}.&-9\leq x< 0\: \: \textrm{atau}\: \: 0<x\leq 9\\ \textrm{d}.&-9< x\leq 9\\ \textrm{e}.&x\in \mathbb{R} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{x^{2}-81}{x^{2}}\geq 0\\ &\displaystyle \frac{(x+9)(x-9)}{x^{2}}\geq 0\\ &\textrm{HP}=\color{red}\left \{ x|x\leq -9\: \: \textrm{atau}\: \: x\geq 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}-4}{x+2}>0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x>2\\ \textrm{b}.&-2\leq x< 2\\ \textrm{c}.&x<-2\: \: \textrm{atau}\: \: x>2\\ \textrm{d}.&x<-2\: \: \textrm{atau}\: \: -2< x< 2\\ \textrm{e}.&x\geq -2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{x^{2}-4}{x+2}>0\\ &\displaystyle \frac{(x+2)(x-2)}{(x+2)}>0\\ &(x-2)>0\\ &\color{red}x>2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 51.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}<0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|-9< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{b}.&\left \{ x|-6< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \}\\ \color{red}\textrm{d}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: \displaystyle \frac{5}{2}<x<5,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\left \{ x|x< -9\: \: \textrm{atau}\: \: -6< x< \displaystyle \frac{5}{2}\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}&<0\\ \displaystyle \frac{(x+6)(x-5)}{(x+9)(2x-5)}&<0\\ \color{red}\textrm{Cukup jelas}& \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 52.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{2x+6}{x-4}\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-10<x<4\\ \color{red}\textrm{b}.&-10\leq x<4\\ \textrm{c}.&-4<x\leq 10\\ \textrm{d}.&x\leq -10\: \: \textrm{atau}\: \: x\geq 4\\ \textrm{e}.&x<-10\: \: \textrm{atau}\: \: x\geq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{2x+6}{x-4}&\leq 1\\ \displaystyle \frac{2x+6}{x-4}-1&\leq 0\\ \displaystyle \frac{2x+6-(x-4)}{x-4}&\leq 0\\ \displaystyle \frac{x+10}{x-4}&\leq 0\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 53.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \displaystyle \frac{x^{2}-x}{x+3}\geq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-3\: \: \textrm{atau}\: \: -1\leq x\leq 3\\ \color{red}\textrm{b}.&-3< x\leq -1\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{c}.&-3\leq x\leq 3\\ \textrm{d}.&-3\leq x\leq -1\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{e}.&-3\leq x\leq -1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{x^{2}-x}{x+3}&\geq 1\\ \displaystyle \frac{x^{2}-x}{x+3}&-1\geq 0\\ \displaystyle \frac{x^{2}-x-(x+3)}{x+3}&\geq 0\\ \displaystyle \frac{x^{2}-2x-3}{x+3}&\geq 0\\ \displaystyle \frac{(x-3)(x+1)}{x+3}&\geq 0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 54.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x+2+\displaystyle \frac{1}{x+4}>0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-4\: \: \textrm{atau}\: \: x\geq -3\\ \textrm{b}.&x<-4\: \: \textrm{atau}\: \: x>-3\\ \textrm{c}.&-4\leq x\leq -3\\ \color{red}\textrm{d}.&x>-4\\ \textrm{e}.&-4\leq x\leq -3\: \: \textrm{atau}\: \: x>-3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}x+2+\displaystyle \frac{1}{x+4}&>0\\ \displaystyle \frac{(x+2)(x+4)+1}{(x+4)}&>0\\ \displaystyle \frac{x^{2}+6x+8+1}{x+4}&>0\\ \displaystyle \frac{x^{2}+6x+9}{x+4}&>0\\ \displaystyle \frac{(x+3)^{2}}{(x+4)}&>0\\ x&>-4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 55.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x+3<\displaystyle \frac{x^{2}+6x+11}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|x< -3\displaystyle \frac{2}{3}\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \}\\ \textrm{b}.&\left \{ x|0\leq x\leq 11,\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\left \{ x|x<-11\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \}\\ \textrm{d}.&\left \{ x|x<0\: \: \textrm{atau}\: \: x>11,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\left \{ x|x\leq 11\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}x+3<\displaystyle \frac{x^{2}+6x+11}{x}&\\ x+3-\displaystyle \frac{x^{2}+6x+11}{x}&<0\\ \displaystyle \frac{x(x+3)-\left (x^{2}+6x+11 \right )}{x}&<0\\ \displaystyle \frac{x^{2}+3x-x^{2}-6x-11}{x}&<0\\ \displaystyle \frac{-3x-11}{x}&<0\\ \displaystyle \frac{3x+11}{x}&>0 \end{aligned} \end{array}$.