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

$\begin{array}{ll}\\ 6.&\textrm{Penyelesaian dari pertidaksamaan}\\ &\left | \displaystyle \frac{3-2x}{-5} \right |>5\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-11\: \: \textrm{atau}\: x>14\\ \textrm{b}.&x<-14\: \: \textrm{atau}\: x>11\\ \textrm{c}.&11<x<14\\ \textrm{d}.&-14<x<-11\\ \textrm{e}.&x>14 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\left | \displaystyle \frac{3-2x}{-5} \right |&>5\\ \displaystyle \frac{3-2x}{-5}<&-5\: \: \textrm{atau}\: \: \displaystyle \frac{3-2x}{-5}>5\\ \displaystyle \frac{2x-3}{5}>&5\: \: \textrm{atau}\: \: \displaystyle \frac{2x-3}{5}<-5\\ 2x-3>&25\: \: \textrm{atau}\: \: 2x-3<-25\\ 2x>25&+3\: \: \textrm{atau}\: \: 2x<-25+3\\ x>14&\: \: \textrm{atau}\: \: x<-11,\\ &\textrm{dapat juga dituliskan}\\ x<-&11\: \: \textrm{atau}\: \: x>14 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 2-2\left | x+1 \right | \right |>4\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-4\: \: \textrm{atau}\: x>2\\ \textrm{b}.&x<-3\: \: \textrm{atau}\: x>1\\ \textrm{c}.&x<-2\: \: \textrm{atau}\: x>0\\ \textrm{d}.&x<-1\: \: \textrm{atau}\: x>3\\ \textrm{e}.&x<0\: \: \textrm{atau}\: x>4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\left | 2-2\left | x+1 \right | \right |>4&\\ 2-2\left | x+1 \right |<-4&\: \: \textrm{atau}\: \: 2-2\left | x+1 \right |>4\\ -2\left | x+1 \right |<-6\: \: &\textrm{atau}\: \: -2\left | x+1 \right |>2\\ \left | x+1 \right |>3\: \: &\textrm{atau}\: \: \left | x+1 \right |<-1\\ \left\{\begin{matrix} (x+1)<-3\\ (x+1)>3 \end{matrix}\right.\: \: &\textrm{atau}\: \: \left\{\begin{matrix} \left | x+1 \right |<-1\\ \color{red}\textbf{tak mungkin} \end{matrix}\right.\\ \textrm{Selanjutnya}&\: \textrm{akan didapatkan}\\ x<-4\: \: &\textrm{atau}\: \: x>2 \end{aligned} \end{array}$

$\begin{array}{l}\\ 8.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 3-\left | x \right | \right |<10\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-14\: \: \textrm{atau}\: x>12\\ \textrm{b}.&x<-13\: \: \textrm{atau}\: x>13\\ \textrm{c}.&x<-12\: \: \textrm{atau}\: x>10\\ \textrm{d}.&0<x<10\\ \color{red}\textrm{e}.&-13<0<13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\left | 3-\left | x \right | \right |&< 10\\ -10< 3-&\left | x \right |< 10\\ -13< -&\left | x \right |< 7\\ -7< &\left | x \right |\leq 13,\: \: (\textrm{ingat harga}\: \: \left | x \right |\geq 0)\\ 0\leq &\left | x \right |< 13\\ &\textrm{selanjutnya},\\ &\left | x \right |< 13\\ -13< &\: x< 13\\ \textrm{HP}=&\color{red}\left \{ x|\: -13< x< 13,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 10.&(\textrm{SPMB 05})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x-2 \right |^{2}<4\left | x-2 \right |+12\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x\in \mathbb{R}|2\leq x\leq 8 \right \}\\ \textrm{b}.&\left \{ x\in \mathbb{R}|4<x< 8 \right \}\\ \color{red}\textrm{c}.&\left \{ x\in \mathbb{R}|-4<x< 8 \right \}\\ \textrm{d}.&\left \{ x\in \mathbb{R}|-2<x<4 \right \}\\ \textrm{e}.&\left \{ x\in \mathbb{R}|2<x<4 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{misalkan}\: \: p&=\left | x-2 \right |,\: \: \textrm{selanjutnya}\\ \left | x-2 \right |^{2}<&\, 4\left | x-2 \right |+12\\ p^{2}<&\, 4p+12\\ p^{2}-&4p-12<0\\ (p-6)&(p+2)<0\\ -2<p&<6,\: \: \color{magenta}\textrm{atau jika dikembalikan}\\ -2<&\left | x-2 \right |<6,\: \: \color{black}\textrm{ingat, nilanya tidak negatif}\\ 0\leq &\left | x-2 \right |<6\\ -6<&\: x-2<6\\ -4<&\: x<8\\ \textrm{HP}=&\color{red}\left \{ -4<x<8,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$