Lanjutan 3 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Wajib

Dalam menentukan batas interval pada contoh soal-soal penyelesaian berikut bisa merujuk di sini dan di sini (untuk yang berkaitan dengan nilai mutlak), moga dapat membantu Anda dalam memahami interval dalam tulisan berikut

\begin{array}{ll}\\ \fbox{22}.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \left | \displaystyle \frac{5}{4x-3} \right |\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{A}.&-\displaystyle \frac{1}{2}\leq x< \frac{3}{4}\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{B}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \frac{3}{4}< x\leq 2\\ \textrm{C}.&-\displaystyle \frac{1}{2}\leq x\leq 2,\: x\neq \frac{3}{4}\\ \textrm{D}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x>\frac{3}{4}\\ \textrm{E}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x\geq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{E}\\ &\begin{aligned}\left | \displaystyle \frac{5}{4x-3} \right |&\leq 1\\ 5&\leq \left | 4x-3 \right |\\ \left | 4x-3 \right |&\geq 5\\ \left | 4x-3 \right |^{2}&\geq 5^{2}\\ (4x-3)^{2}-5^{2}&\geq 0\\ (4x-3+5)(4x-3-5)&\geq 0\\ (4x+2)(4x-8)&\geq 0\\ 8(2x+1)(x-2)&\geq 0\\ (2x+1)(x-2)&\geq 0\\ \therefore \: x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x&\geq 2 \end{aligned} \end{array}.
\begin{array}{ll}\\ \fbox{23}.&\textrm{(SPMB2002)Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: x\left | x \right |-2< x-2\left | x \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{A}.&x<-2\: \: \textrm{atau}\: \: -1<x<1\\ \textrm{B}.&-2<x<-1\: \: \textrm{atau}\: \: x>2\\ \textrm{C}.&-2<x<1\: \: \textrm{atau}\: \: x>2\\ \textrm{D}.&-1<x<1\: \: \textrm{atau}\: \: x>2\\ \textrm{E}.&-1<x<2\: \: \textrm{atau}\: \: x>3 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}x\left | x \right |-2< x-2\left | x \right |&\\ x\left | x \right |+2\left | x \right |-x-2&< 0\\ (x+2)\left | x \right |-(x+2)&<0\\ (x+2)\left ( \left | x \right |-1 \right )&<0,\: \: \textrm{atau}\: \: \left ( \left | x \right |-1 \right )(x+2)<0\\ \textrm{Kemungkinan pertama}&\: \textrm{yaitu}:\\ (x+2)>0\: \: \textrm{atau}\: \: \left | x \right |-1&<0\\ x>-2\: \: \textrm{atau}\: \: \left | x \right |&<1\Rightarrow -1<x<1\\ \textrm{Kemungkinan yang ke}&\textrm{dua},\\ \left | x \right |-1>0\: \: \textrm{atau}\: \: (x+2)&<0\\ \left | x \right |>1\: \: \textrm{atau}\: \: x<&-2\: \Rightarrow x<-2\\ \therefore ,\: x<-2\: \: \textrm{atau}\: \: -1<x&<1 \end{aligned} \end{array}

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