Pertidaksaan (Lanjutan)

Pertidaksamaan Bentuk Akar

\textbf{Bentuk Umum}:\: \: \sqrt{f(x)}\cdots \sqrt{g(x)}\: \begin{cases} \sqrt{f(x)}< \sqrt{g(x)}\\\\ \sqrt{f(x)}\leq \sqrt{g(x)} \\\\ \sqrt{f(x)}> \sqrt{g(x)} \\\\ \sqrt{f(x)}\geq \sqrt{g(x)} \end{cases}.

Dalam mengerjakan permasalahan pertidaksamaan bentuk akar (irasional) adalah :

  1. Mengkuadratkan masing-masing ruas.
  2. Di bawah tanda akar (syarat numerus) adalah  \geq .
  3. Himpunan penyelesaian merupakan irisan penyelesaian yang ada.

\LARGE{\fbox{\LARGE{\fbox{Contoh Soal}}}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Tentukanlah himpunan penyelesaian dari pertidaksamaan bentuk akar}\\ &\begin{array}{lllllllll}\\ \textrm{a}.\quad \sqrt{x}< 1&\textrm{d}.\quad \sqrt{x}< -1\\ \textrm{b}.\quad \sqrt{x-1}< 2&\textrm{e}.\quad \sqrt{x-1}< -2\\ c.\quad \sqrt{x+1}< 3 &\textrm{f}.\quad \sqrt{x+1}< -3\end{array} \end{array}.

Jawab:

\textbf{a}.\quad \textrm{Untuk}\: \: \sqrt{x}< 1\\ \textrm{(i)}\quad \textrm{Syarat numerus}:x\geq 0\\ \textrm{(ii)}\quad x< 1\qquad \textrm{(dikuadratkan masing-masing ruas)}\\ \textrm{(iii)}\: \: \textrm{HP}=\left \{ x|\: \: 0\leq x< 1,\: x\in \mathbb{R} \right \}\\ \textrm{(iv)}\: \: \textrm{Garis bilangan}.

336

\textbf{b}.\quad \textrm{Untuk}\: \: \sqrt{x-1}< 2\\ \textrm{(i)}\quad \textrm{Syarat numerus}:x-1\geq 0\: \: \Rightarrow \: \: x\geq 1\\ \textrm{(ii)}\quad x-1< 4\qquad \textrm{(dikuadratkan masing-masing ruas)}\: \: \Rightarrow \: \: x< 5\\ \textrm{(iii)}\: \: \textrm{HP}=\left \{ x|\: \: 1\leq x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{(iv)}\: \: \textrm{Garis bilangan}.

337

\textbf{d}.\quad \textrm{Untuk}\: \: \sqrt{x}< -1\\ \textrm{(i)}\quad \textrm{Syarat numerus}:x\geq 0\quad \textrm{(tidak ada bilangan real yang memenuhi)}\\ \textrm{(ii)}\: \: \: \textrm{HP}=\left \{ \right \}.

Pertidaksaan Bentuk Harga Mutlak

Untuk bilangan real x secara definisi nilai mutlak disimbolkan dengan  \left | x \right |.

dengan

\left | x \right |=\begin{cases} x,\quad \textrm{untuk}\: \: x> 0\\ 0,\quad \textrm{untuk}\: \: x=0 \\ -x,\quad \textrm{untuk}\: \: x< 0 \end{cases}.

Beberapa hal yang perlu diperhatikan dalam menyelesaikan pertidaksamaan harga mutlak

\begin{array}{ll}\\ \square &\textrm{untuk}\: a \: \textrm{dan}\: b\: \textrm{bilangan real, berlaku}\\ 1.&\sqrt{a^{2}}=\left | a \right |\: \: \Leftrightarrow \: \: a^{2}=\left | a \right |^{2}\\ 2.&\left | a-b \right |=\begin{cases} a-b ,\: \: \textrm{jika}\: \: a\geq b \\ -(a-b)=b-a\: \: \textrm{jika}\: \: a< b \end{cases}\\ 3.&\left | a-b \right |=\left | b-a \right |\\ 4.&\left | a.b \right |=\left | a \right |.\left | b \right |\\ 5.&\left | \displaystyle \frac{a}{b} \right |=\displaystyle \frac{\left | a \right |}{\left | b \right |},\: \: \textrm{dengan}\: \: b\neq 0\\ 6.&\left | a \right |-\left | b \right |\leq \left | a+b \right |\leq \left | a \right |+\left | b \right |\\ 7.&\left | \displaystyle \frac{a}{b} \right |< c\: \: \Leftrightarrow \: \: \left | a \right |< c\left | b \right |\: \: \Leftrightarrow \: \: (a+cb).(a-cb)< 0\: \: \: (\textrm{di sini} \: \: c> 0)\\ 8.&\left | \displaystyle \frac{a}{b} \right |> c\: \: \Leftrightarrow \: \: \left | a \right |> c\left | b \right |\: \: \Leftrightarrow \: \: (a+cb).(a-cb)> 0\: \: \: (\textrm{di sini} \: \: c> 0)\\ 9.&\left | f(x) \right |< \left | g(x) \right |\: \: \textrm{dapat diubah menjadi}\: \: f^{2}(x)< g^{2}(x)\\ 10.&\left | f(x) \right |> \left | g(x) \right |\: \: \textrm{dapat diubah menjadi}\: \: f^{2}(x)> g^{2}(x) \end{array}.

\LARGE{\fbox{\LARGE{\fbox{Contoh Soal}}}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Tentukanlah himpunan penyelesaian yang memenuhi}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \left | 2x-7 \right |=3&\textrm{d}.\quad \left | 2x-7 \right |< 3&\textrm{g}.\quad \left | 2x-7 \right |\leq -3\\ \textrm{b}.\quad \left | 2x-7 \right |=-3&\textrm{e}.\quad \left | 2x-7 \right |\geq 3&\textrm{h}.\quad \left | 2x-7 \right |\geq -3\\ \textrm{c}.\quad \left | x-3 \right |=\left | 2-x \right |&\textrm{f}.\quad \left | x-3 \right |\leq \left | 2-x \right |&\textrm{i}.\quad \left | x^{2}-3x+1 \right |\leq 1 \end{array} \end{array}.

Jawab:

\begin{array}{|l|l|l|}\hline \begin{aligned}\textrm{a}.\quad&\left | 2x-7 \right |=3\\ & 2x-7 =\begin{cases} 3 \\ -3 \end{cases}\\ &2x-7=3\: \: \textrm{atau}\: \: 2x-7=-3\\ &2x=3+7\: \: \textrm{atau}\: \: 2x=-3+7\\ &x=\displaystyle \frac{10}{2}=5\: \: \textrm{atau}\: \: x=\displaystyle \frac{4}{2}=2\\ \textrm{HP}&=\left \{ 2,5 \right \}\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad&\left | 2x-7 \right |=-3\\ &\textrm{nilai mutlak}\\ &\textrm{suatu bilangan}\\ &\textrm{tidak mungkin}\\ &\textrm{negatif}\\ &\\ \textrm{HP}&=\left \{ \right \}\\ & \end{aligned}&\begin{aligned}\textrm{c}.\quad&\left | x-3 \right |=\left | 2-x \right |\\ &x-3=\begin{cases} (2-x)\\ -(2-x) \end{cases}\\ &x-3=2-x\: \: \textrm{atau}\: \: x-3=-(2-x)\\ &2x=5\: \: \textrm{atau}\: \: -3=-2\: (\textrm{tidak mungkin})\\ &x=\displaystyle \frac{5}{2}\\ \textrm{HP}&=\left \{ \displaystyle \frac{5}{2} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad&\left | 2x-7 \right |< 3\\ &\textrm{maka}\\ &-3< 2x-7< 3\\ &-3+7< 2x-7+7<3+7\\ &4<2x<10\\ &2<x<5\\ \textrm{HP}&=\left \{ x|\: 2<x<5,\: x\in \mathbb{R} \right \}\\ &\\ & \end{aligned}&\begin{aligned}\textrm{e}.\quad&\left | 2x-7 \right |\geq 3\\ &\textrm{maka}\\ &2x-7\leq -3\: \: \textrm{atau}\: \: 2x-7\geq 3\\ &2x\leq -3+7\: \: \textrm{atau}\: \: 2x\geq 3+7\\ &2x\leq 4\: \: \textrm{atau}\: \: 2x\geq 10\\ &x\leq 2\: \: \textrm{atau}\: \: x\geq 5\\ \textrm{HP}&=\left \{ x|\: x\leq 2\: \: \textrm{atau}\: \: x\geq 5,\: x\in \mathbb{R} \right \}\\ &\\ & \end{aligned}&\begin{aligned}\textrm{f}.\quad&\left | x-3 \right |\leq \left | 2-x \right |\\ &\textrm{dikuadratkan kedua ruas}\\ &\left | x-3 \right |^{2}\leq \left | 2-x \right |^{2}\\ &x^{2}-6x+9\leq 4-4x+x^{2}\\ &-6x+4x\leq 4-9\\ &-2x\leq -5\\ &2x\geq 5\\ &x\geq \displaystyle \frac{5}{2}\\ \textrm{HP}&=\left \{ x|\: x\geq \displaystyle \frac{5}{2},\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Tentukanlah himpunan penyelesaian yang memenuhi}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \left | x-3 \right |=2-x\\ \textrm{b}.\quad \left | x-3 \right |=\left | 2-x \right |\\ \textrm{c}.\quad \left | x-3 \right |\leq \left | 2-x \right |\end{array} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|l|}\hline \begin{aligned}\left | x-3 \right |&=2-x\\ &\textrm{berarti:}\quad &2-x\geq 0\\ &\textrm{sehingga}&x\leq 2\\ \left | x-3 \right |&=\begin{cases} (2-x) \\ -(2-x) \end{cases}\\ x-3&=2-x\: \: \textrm{atau}\: \: x-3=-(2-x)\\ 2x&=5\: \: \textrm{atau}\: \: -3=-2 (\textrm{tak mungkin})\\ x&=\displaystyle \frac{5}{2},\qquad \textrm{padahal}\: \: x\leq 2\\ \textrm{HP}&=\left \{ \right \} \end{aligned}&\begin{aligned}&\textrm{Lihat jawaban/pembahasan}\\ &\textrm{pada No. 1. c} \end{aligned}&\begin{aligned}&\textrm{Lihat jawaban/pembahasan}\\ &\textrm{pada No. 1. f} \end{aligned}\\\hline \end{array}

\begin{array}{ll}\\ \fbox{3}.&\textrm{Tentukanlah himpunan penyelesaian yang memenuhi}\\ &\begin{array}{lll}\\ \textrm{a}.\quad x^{2}-2\left | x \right |-8=0\\ \textrm{b}.\quad x^{2}-2\left | x \right |-8> 0\\ \textrm{c}.\quad x^{2}-2\left | x \right |-8\geq 0\\ \textrm{d}.\quad x^{2}-2\left | x \right |-8< 0\\ \textrm{e}.\quad x^{2}-2\left | x \right |-8\leq 0\end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad&x^{2}-2\left | x \right |-8=0\\ &\left | x \right |^{2}-2\left | x \right |-8=0\\ &\left ( \left | x \right |-4 \right )\left ( \left | x \right |+2 \right )=0\\ &\left | x \right |=4\: \: \textrm{atau}\: \: \left | x \right |=-2\: (\textrm{tidak mungkin})\\ &\left | x \right |=\begin{cases} 4 \\ -4 \end{cases}\\ \textrm{HP}&=\left \{ -4,4 \right \}\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad&x^{2}-2\left | x \right |-8> 0\\ &\blacklozenge \: \: (\textrm{untuk}\: \: x\geq 0,\: \: \textrm{maka}\: \: \left | x \right |=x)\\ &x^{2}-2x-8> 0\\ &\left ( x-4 \right )\left ( x+2 \right )> 0\\ \textrm{A}&=\left \{ x|\: x< -2\: \: \textrm{atau}\: \: x> 4,\: x\in \mathbb{R} \right \}\\ &\textrm{karena}\: \: x\geq 0,\: \: \textrm{maka}\\ \textrm{HP}_{1}&=\left \{ x|\: x> 4,\: x\in \mathbb{R} \right \}\\ &\blacklozenge \: \: (\textrm{untuk}\: \: x< 0,\: \: \textrm{maka}\: \: \left | x \right |=-x)\\ &x^{2}+2x-8> 0\\ &\left ( x+4 \right )\left ( x-2 \right )> 0\\ \textrm{B}&=\left \{ x|\: x<-4\: \: \textrm{atau}\: \: x>2,\: x\in \mathbb{R} \right \}\\ &\textrm{karena}\: \: x< 0,\: \: \textrm{maka}\\ \textrm{HP}_{2}&=\left \{ x|\: x< -4 \right \}\\ &\\ \textrm{Jadi}&,\\ \textrm{HP}_{\left (1\: \cup \: 2 \right )}&=\left \{ x|\: x< -4\: \: \textrm{atau}\: \: x> 4,\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Tentukanlah himpunan penyelesaian yang memenuhi}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \left | 3-\left | x \right | \right |=10\\ \textrm{b}.\quad \left | 3-\left | x \right | \right |> 10\\ \textrm{c}.\quad \left | 3-\left | x \right | \right |\geq 10\\ \textrm{d}.\quad \left | 3-\left | x \right | \right |< 10\\ \textrm{e}.\quad \left | 3-\left | x \right | \right |\leq 10\end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad&\left | 3-\left | x \right | \right |=10\\ &3-\left | x \right |=\begin{cases} 10 \\ -10 \end{cases}\\ &3-\left | x \right |=10\: \: \textrm{atau}\: \: 3-\left | x \right |=-10\\ &\left | x \right |=-7\: \: (\textrm{tidak mungkin})\: \: \textrm{atau}\: \: \left | x \right |=13\\ \textrm{HP}&=\left \{ -13,13 \right \}\\ &\\ &\\ &\end{aligned}&\begin{aligned}\textrm{d}.\quad&\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{sehingga},\\ &\left | x \right |< 13\\ &-13< x< 13\\ \textrm{HP}&=\left \{ x|\: -13< x< 13,\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Himpunan penyelesaian yang memenuhi}\: \: \left | \displaystyle \frac{3-2x}{2+x} \right |\leq 4\\ &\begin{array}{lll}\\ \end{array} \end{array}\\\\ \textrm{Jawab}:\\ \textrm{Perhatikan bahwa soal bentuk di atas memenuhi}\: \: \left | \displaystyle \frac{a}{b} \right |\leq c,\: \: \textrm{dengan}\: \: \: \begin{cases} a&= 3-2x \\ b&=2+x \\ c&= 4 \end{cases}\\ \begin{aligned}\left ( \left (3-2x \right )+4(2+x) \right )\left ( \left (3-2x \right )-4(2+x) \right )&\leq 0\\ (2x+11)(-6x-5)&\leq 0\\ (2x+11)(6x+5)&\geq 0\\ x=-\displaystyle \frac{11}{2}\: \: \textrm{atau}\: \: x=\displaystyle &-\frac{5}{6}\\ \textrm{HP}=\left \{ x|\: x\leq -\displaystyle \frac{11}{2}\: \: \textrm{atau}\: \: x\geq \displaystyle -\frac{5}{6},\: x\in \mathbb{R} \right \}& \end{aligned}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Nilai-nilai x yang memenuhi}\: \: \left | \displaystyle x+5 \right |>\displaystyle \frac{x-1}{\left | x+1 \right |}\: \: \textrm{adalah ....} \\ &\begin{array}{ll}\\ \textrm{a}.&x<-6\: \: \textrm{atau}\: \: x>1\\ \textrm{b}.&-6<x<-4\\ \textrm{c}.&-\infty < x< -1\: \: \textrm{atau}\: \:-1<x<\infty \\ \textrm{d}.&x>1\\ \textrm{e}.&\textrm{tidak ada nilai x yang memenuhi} \end{array} \\\end{array}.

Jawab: C

Sebagai pengingat x ≠ – 1.

Perhatikan bahwa

\underset{\overbrace{-\infty\qquad\qquad \underset{-x-5\: \: \Leftarrow =\: \: \overbrace{ \left | x+5 \right | }\: \: =\Rightarrow \: \: x+5}{\underbrace{-5}}\qquad\qquad \underset{-x-1\: \: \Leftarrow =\: \: \overbrace{ \left | x+1 \right | }\: \: =\Rightarrow \: \: x+1}{\underbrace{-1}}\qquad\qquad \infty }}{\underbrace{\left | \displaystyle x+5 \right |>\displaystyle \frac{x-1}{\left | x+1 \right |}}}.

\begin{array}{|c|c|c|}\hline \begin{aligned}\textrm{untuk}&\\ &-\infty <x< -5 \end{aligned}&\begin{aligned}\textrm{untuk}&\\ &-5\leq x< -1 \end{aligned}&\begin{aligned}\textrm{untuk}&\\ &-1<x<\infty \end{aligned}\\ \begin{cases} \left | x+5 \right |=-(x+5) \\ \left | x+1 \right |=-(x+1) \end{cases}&\begin{cases} \left | x+5 \right |=(x+5) \\ \left | x+1 \right |=-(x+1) \end{cases}&\begin{cases} \left | x+5 \right |=(x+5) \\ \left | x+1 \right |=(x+1) \end{cases}\\\hline \textbf{Kasus Pertama}&\textbf{Kasus kedua}&\textbf{Kasus ketiga}\\\hline \multicolumn{3}{c}{.}\\ \multicolumn{3}{c}{\textrm{selanjutnya perhatikanlah uraian berikut}} \end{array}.

Kasus Pertama

untuk  x< -5.

\begin{aligned}\left | \displaystyle x+5 \right |&>\displaystyle \frac{x-1}{\left | x+1 \right |}\\ -(x+5)&>\displaystyle \frac{x-1}{-(x+1)}\\ -(x+5)+\displaystyle \frac{x-1}{x+1}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{-(x+5)(x+1)+x-1}{(x+1)}&>0\\ \displaystyle \frac{-x^{2}-5x-6}{(x+1)}&>0\\ \displaystyle \frac{x^{2}+5x+6}{(x+1)}&<0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+2)(x+3)}{(x+1)}&<0\\ \textrm{A}=\left \{ x|\: x<-3\: \textrm{atau}\: -2<x<-1,\: x\in \mathbb{R} \right \}&\\ \textrm{karena}\: \: x<-5&\\ \textrm{HP}_{1}=\left \{ x|\: x<-5 \right \} \end{aligned}.

Kasus kedua

Untuk  -5\leq x< -1.

\begin{aligned}\left | \displaystyle x+5 \right |&>\displaystyle \frac{x-1}{\left | x+1 \right |}\\ (x+5)&>\displaystyle \frac{x-1}{-(x+1)}\\ (x+5)+\displaystyle \frac{x-1}{x+1}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+5)(x+1)+x-1}{(x+1)}&>0\\ \displaystyle \frac{x^{2}-5x-6}{(x+1)}&>0\\ \displaystyle \frac{x^{2}+7x+4}{(x+1)}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+\displaystyle \frac{7+\sqrt{33}}{2})(x+\displaystyle \frac{7-\sqrt{33}}{2})}{(x+1)}&>0\\ \textrm{B}=\left \{ x|\: x<\displaystyle \frac{-7-\sqrt{33}}{2}\: \textrm{atau}\: x>\displaystyle \frac{-7+\sqrt{33}}{2},\: x\in \mathbb{R} \right \}&\\ \textrm{karena}\: \: -5\leq x<-1&\\ \textrm{HP}_{2}=\left \{ x|\: -5\leq x<-1 \right \}& \end{aligned}.

Kasus ketiga

Untuk  x>-1.

\begin{aligned}\left | \displaystyle x+5 \right |&>\displaystyle \frac{x-1}{\left | x+1 \right |}\\ (x+5)&>\displaystyle \frac{x-1}{(x+1)}\\ (x+5)-\displaystyle \frac{x-1}{x+1}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+5)(x+1)-x+1}{(x+1)}&>0\\ \displaystyle \frac{x^{2}+5x+6}{(x+1)}&>0\\ \displaystyle \frac{(x+2)(x+3)}{(x+1)}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+2)(x+3)}{(x+1)}&>0\\ \textrm{C}=\left \{ x|\: -3<x<-2\: \textrm{atau}\: x>-1,\: x\in \mathbb{R} \right \}&\\ \textrm{karena}\: \: x>-1&\\ \textrm{HP}_{3}=\left \{ x|\: x>-1 \right \}& \end{aligned}.

Jadi,  \textrm{HP}=\textrm{HP}_{1}\: \: \cup \: \: \textrm{HP}_{2}\: \: \cup \: \: \textrm{HP}_{3}=\left \{ x|\: x<-1\: \: \textrm{atau}\: \: x>-1,\: x\in \mathbb{R} \right \}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Jika}\: \: 3<x<5\: \: \textrm{maka penyelesaian untuk} \\ &\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{ll}\\ \textrm{a}.&2x-2\\ \textrm{b}.&2\\ \textrm{c}.&8-2x\\ \textrm{d}.&-2\\ \textrm{e}.&2x-8 \end{array} \\\end{array}\\\\\\ \textrm{Jawab}:\: \textbf{E}\\\\ \begin{aligned}\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}&=\sqrt{(x-3)^{2}}-\sqrt{(x-5)^{2}}=\left | x-3 \right |-\left | x-5 \right |\\ \textrm{ingat bahwa saat}\: \: 3<x<5\: \: &\textrm{maka}\: \begin{cases} \left | x-3 \right |=(x-3) \\ \left | x-5 \right |=-(x-5) \end{cases},\: \textrm{sehingga}\\ &=\left | x-3 \right |-\left | x-5 \right |=(x-3)-\left ( -(x-5) \right )\\ &=x-3+x-5\\ &=2x-8 \end{aligned}.

\LARGE{\fbox{\LARGE{\fbox{Latihan Soal}}}}.

  1. Kerjakanlah soal pada contoh soal N0.1 yang belum jawab atau dibahas.
  2. Kerjakan juga soal pada contoh soal N0.3 yang belum jawab atau dibahas.
  3. Kerjakan juga soal pada contoh soal N0.4 yang belum jawab atau dibahas.
  4. Tentukanlah nilai  x  yang memenuhi

\begin{array}{lllll}\\ .\quad&\textrm{a}.\quad \left | 3x-2 \right |\geq 4&\textrm{f}.\quad 2\left | x-1 \right |< \left | x+2 \right |&\textrm{k}.\quad \left | x-1 \right |^{2}+2\left | x-1 \right |< 15&\textrm{p}.\quad \left | \displaystyle \frac{x-1}{x+2} \right |\leq \left | \displaystyle \frac{x+3}{x-4} \right |\\ &\textrm{b}.\quad \left | 2x-3 \right |< 7&\textrm{g}.\quad \left | \displaystyle \frac{1}{4}x^{2}-10 \right |< 6&\textrm{l}.\quad \left | x-3 \right |^{2}> 4\left | x-3 \right |+12&\textrm{q}.\quad \left | \displaystyle \frac{3}{2x-1} \right |> 1\\ &\textrm{c}.\quad \left | 3x+2 \right |> 5&\textrm{h}.\quad \left | \left | x \right |+x \right |\leq 2&\textrm{m}.\quad \left | 2x+1 \right |+6< 5\left | 2x+1 \right |^{2}&\textrm{r}.\quad x+\left | x-3 \right |\leq 3\\ &\textrm{d}.\quad 0< \left | x-3 \right |\leq 3&\textrm{i}.\quad \left | 3x-\left | x \right | \right |\geq 2&\textrm{n}.\quad \left | \displaystyle \frac{2x+1}{x-2} \right |< 1&\textrm{s}.\quad \left | x-2 \right |^{2}-6+2x< 0\\ &\textrm{e}.\quad \left | 2x+1 \right |< \left | 2x-3 \right |&\textrm{j}.\quad \left | 9-2x \right |> \left | 4x \right |&\textrm{o}.\quad \left | \displaystyle \frac{3x-2}{x+1} \right |> 2 \end{array}.

Sumber Referensi

  1. Marwanta, dkk. 2013. Matematika SMA Kelas X. Jakarta: Yudistira
  2. Marwanta, Sigit Suprijanto, Herynugroho, Suwarni Murniati, Kamta Agus Sajak, Soetiyono. 2004. Matematika Interaktif 1A Kelas 1 SMA Semester Pertama. Jakarta: Yudistira.
  3. Sukino.  2013. Matematika Kelompok Peminatan Matematika dan Ilmu Alam untuk SMA/MA Kelas x. Jakarta: Erlangga.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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