Contoh Soal 2 Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 6.& \text{Penyelesaian dari pertidaksamaan}\\ & \left|{\displaystyle \frac{3-2x}{-5}}\right|>5\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-11\text{atau}x>14\\ \text{b}.& x<-14\text{atau}x>11\\ \text{c}.& 11<x<14\\ \text{d}.& -14<x<-11\\ \text{e}.& x>14\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{3-2x}{-5}}\right|& >5\\ {\displaystyle \frac{3-2x}{-5}<}& -5\text{atau}{\displaystyle \frac{3-2x}{-5}>5}\\ {\displaystyle \frac{2x-3}{5}>}& 5\text{atau}{\displaystyle \frac{2x-3}{5}<-5}\\ 2x-3>& 25\text{atau}2x-3<-25\\ 2x>25& +3\text{atau}2x<-25+3\\ x>14& \text{atau}x<-11,\\ & \text{dapat juga dituliskan}\\ x<-& 11\text{atau}x>14\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |2-2|x+1||>4\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-4\text{atau}x>2\\ \text{b}.& x<-3\text{atau}x>1\\ \text{c}.& x<-2\text{atau}x>0\\ \text{d}.& x<-1\text{atau}x>3\\ \text{e}.& x<0\text{atau}x>4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}|2-2|x+1||>4& \\ 2-2|x+1|<-4& \text{atau}2-2|x+1|>4\\ -2|x+1|<-6& \text{atau}-2|x+1|>2\\ |x+1|>3& \text{atau}|x+1|<-1\\ \{\begin{array}{c}(x+1)<-3\\ (x+1)>3\end{array}& \text{atau}\{\begin{array}{c}|x+1|<-1\\ \mathbf{\text{tak mungkin}}\end{array}\\ \text{Selanjutnya}& \text{akan didapatkan}\\ x<-4& \text{atau}x>2\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |3-\left|x\right||<10\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-14\text{atau}x>12\\ \text{b}.& x<-13\text{atau}x>13\\ \text{c}.& x<-12\text{atau}x>10\\ \text{d}.& 0<x<10\\ \text{e}.& -13<0<13\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}|3-\left|x\right||& <10\\ -10<3-& \left|x\right|<10\\ -13<-& \left|x\right|<7\\ -7<& \left|x\right|\le 13,(\text{ingat harga}\left|x\right|\ge 0)\\ 0\le & \left|x\right|<13\\ & \text{selanjutnya},\\ & \left|x\right|<13\\ -13<& x<13\\ \text{HP}=& \{x|-13<x<13,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{(UM UGM 05)}\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & |x^2-3|<2x\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -1<x<3\\ \text{b}.& -3<x<1\\ \text{c}.& 1<x<3\\ \text{d}.& -3<x<-1\text{atau}1<x<3\\ \text{e}.& x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}|x^2-3|& <2x\\ -2x<(x^2-3)& <2x\\ \text{dipartisi men}& \text{jadi dua bagian}\\ \bullet \text{pertama}& \\ (x^2-3)& >-2x\\ x^2+2x-3& >0\\ (x+3)(x-1)& >0\\ x<-3\text{atau}& x>1\\ \bullet \text{kedua}& \\ (x^2-3)& <2x\\ x^2-2x-3& <0\\ (x-3)(x+1)& <0\\ -1<x<3& \\ \text{ambil yang}& \text{memenuhi keduanya}\\ \text{berupa iris}& \text{an}\\ \text{HP}=& \{1<x<3,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& (\text{SPMB 05})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {|x-2|}^2<4|x-2|+12\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \{x\in \mathbb{R}|2\le x\le 8\}\\ \text{b}.& \{x\in \mathbb{R}|4<x<8\}\\ \text{c}.& \{x\in \mathbb{R}|-4<x<8\}\\ \text{d}.& \{x\in \mathbb{R}|-2<x<4\}\\ \text{e}.& \{x\in \mathbb{R}|2<x<4\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{misalkan}p& =|x-2|,\text{selanjutnya}\\ {|x-2|}^2<& 4|x-2|+12\\ p^2<& 4p+12\\ p^2-& 4p-12<0\\ (p-6)& (p+2)<0\\ -2<p& <6,\text{atau jika dikembalikan}\\ -2<& |x-2|<6,\text{ingat, nilanya tidak negatif}\\ 0\le & |x-2|<6\\ -6<& x-2<6\\ -4<& x<8\\ \text{HP}=& \{-4<x<8,x\in \mathbb{R}\}\end{array}\end{array}$