MTK Wajib

Persamaan dan Pertidaksamaan Nilai Mutlak Linear Satu Variable

By Nadya Putri Safira - Maret 07, 2019

A. Nilai Mutlak
Nilai mutlak suatu bilangan adalah adalah jarak suatu bilangan yang diukur dari posisi Odari bilangan itu pada garis bilangan real.
Perhatikanlah gambar berikut sebagai ilustrasi
Selanjutnya untuk setiap bilangan real  x  berlaku:
\begin{aligned}&\\ \left | x \right |&=\begin{cases} x & \text{ jika } x\geq 0 \\\\ -x & \text{ jika } x< 0 \end{cases}\\ & \end{aligned}.
\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.
\begin{array}{llll}\\ 1&\textrm{a}.&\left | 1 \right |=1\\ &\textrm{b}.&\left | -1 \right |=1&,\textrm{ingat jarak bilangan} \: \: -1\: \: \textrm{terhadap} \: \: 0 \\ &\textrm{c}.&\left | 2 \right |=2\\ &\textrm{d}.&\left | -2 \right |=2&,\textrm{ingat jarak bilangan} \: \: -2\: \: \textrm{terhadap} \: \: 0\\ &\textrm{e}.&\left | \displaystyle \frac{2}{3} \right |=\displaystyle \frac{2}{3}\\ &\textrm{f}.&\left | -\displaystyle \frac{2}{3} \right |=\displaystyle \frac{2}{3}&,\textrm{ingat jarak bilangan} \: \: -\displaystyle \frac{2}{3}\: \: \textrm{terhadap} \: \: 0\\ &\textrm{g}.&\left | 1+\sqrt{5} \right |=1+\sqrt{5}\\ &\textrm{h}.&\left | -1-\sqrt{5} \right |=1+\sqrt{5}\\ &\textrm{i}.&\left | 1-\sqrt{5} \right |=\sqrt{5}-1&,\textrm{ingat jarak bilangan} \: \: 1-\sqrt{5}\: \: \textrm{terhadap} \: \: 0\\ &\textrm{j}.&\left | -1+\sqrt{5} \right |=\sqrt{5}-1\\\\ 2.&\textrm{a}.&\left | 1 \right |+\left | -1 \right |+\left | 2 \right |+\left | -2 \right |\\ &&=1+1+2+2\\ &&=6\\ &\textrm{b}.&\left | 1+\sqrt{5} \right |+\left |1-\sqrt{5} \right |&\\ &&=1+\sqrt{5}+\sqrt{5}-1&\\ &&=2\sqrt{5}\end{array}.
\begin{array}{llll}\\ 3&\textrm{a}.&\left | 1-\left | -2 \right | \right |+\left | 2-\left | -3 \right | \right |+\left | 3-\left | -4 \right | \right |+\left | 4-\left | -5 \right | \right |\\ &&=\left | 1-2 \right |+\left | 2-3 \right |+\left | 3-4 \right |+\left | 4-5 \right |\\ &&=\left | -1 \right |+\left | -1 \right |+\left | -1 \right |+\left | -1 \right |\\ &&=1+1+1+1\\ &&=4\\ &\textrm{b}.&\left | \displaystyle \frac{1}{1-\sqrt{2}} \right |+\left |\displaystyle \frac{1}{\sqrt{2}-\sqrt{3}} \right |+\left |\displaystyle \frac{1}{\sqrt{3}-2} \right |\\ &&=\left | \displaystyle \frac{1}{1-\sqrt{2}}\times \frac{1+\sqrt{2}}{1+\sqrt{2}} \right |+\left |\displaystyle \frac{1}{\sqrt{2}-\sqrt{3}}\times \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}} \right |+\left |\displaystyle \frac{1}{\sqrt{3}-2}\times \frac{\sqrt{3}+2}{\sqrt{3}+2} \right |\\ &&=\left | -\left ( 1+\sqrt{2} \right ) \right |+\left | -\left ( \sqrt{2}+\sqrt{3} \right ) \right |+\left | -\left ( \sqrt{3}+2 \right ) \right |\\ &&=\left ( 1+\sqrt{2} \right )+\left ( \sqrt{2}+\sqrt{3} \right )+\left ( \sqrt{3}+2 \right )\\ &&=3+2\left ( \sqrt{2}+\sqrt{3} \right ) \end{array}.
B. Persamaan Nilai Mutlak Linear Satu Variabel
Perhatikanlah ilustrasi untuk persamaan  y=\left | x \right |   berikut ini
Persamaan nilai multak adalah sebuah kalimat yang memuat variabel berderajat(pangkat) satu dalam tanda nilai mutlak (“|…|”) dan dihubungkan dengan tanda sama dengan.
\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.
\begin{array}{llll}\\ \multicolumn{4}{l}{\textrm{Selesaikanlah soal berikut ini!}}\\ \textrm{a}.&x=\left | 2 \right |&\textrm{h}.&-2\left | x-3 \right |+5=7\\ \textrm{b}.&\left | 2x \right |=\left | -5 \right |&\textrm{i}.&-2\left | -(x-3) \right |+\left | x-3 \right |+5=7\\ \textrm{c}.&\left | 2x-1 \right |=0&\textrm{j}.&\left | x-2 \right |=\left | x+1 \right |\\ \textrm{d}.&\left | 2x-1 \right |=7&\textrm{k}.&\left | 2-x \right |=\left | x+1 \right |\\ \textrm{e}.&\left | 2x-1 \right |=-7&\textrm{l}.&\left | 2x-1 \right |=\left | x+1 \right |\\ \textrm{f}.&\left | x-3 \right |+5=7&\\ \textrm{g}.&2\left | x-3 \right |+5=7& \end{array}.
Jawab:
\begin{array}{|c|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}&\textrm{d}\\\hline \begin{aligned}x&=\left | 2 \right |\\ &=2\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\left | 2x \right |&=\left | -5 \right |\\ \left | 2x \right |&=\left | 5 \right |\\ (2x)&=\pm 5\\ &\begin{cases} 2x &=+5 \\ \qquad x&=\displaystyle \frac{5}{2}\\ \textrm{atau}&\\ 2x &=-5\\ \qquad x&=-\displaystyle \frac{5}{2} \end{cases}\\ & \end{aligned}&\begin{aligned}\left | 2x-1 \right |&=0\\ (2x-1)&=\pm 0\\ (2x-1)&=0\\ 2x&=1\\ x&=\displaystyle \frac{1}{2}\\ &\\ &\\ &\\ &\\ \end{aligned}&\begin{aligned}\left | 2x-1 \right |&=7\\ (2x-1)&=\pm 7\\ 2x&=1\pm 7\\ x&=\displaystyle \frac{1\pm 7}{2}\\ &\begin{cases} x &=\displaystyle \frac{1+7}{2} \\ &=4\\ \textrm{atau}&\\ x &=\displaystyle \frac{1-7}{2}\\ &=-3 \end{cases} \end{aligned}\\\hline \textrm{e}&\textrm{f}&\textrm{g}&\textrm{h}\\\hline \begin{aligned}\left | 2x-1 \right |&=-7\\ &\\ \textrm{tidak}\: &\: \textrm{ada solusi}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\left | x-3 \right |+5&=7\\ \left | x-3 \right |&=7-5\\ \left | x-3 \right |&=2\\ (x-3)&=\pm 2\\ x&=3\pm 2\\ &\begin{cases} x & =5 \\ x & =1 \end{cases} \end{aligned}&\begin{aligned}2\left | x-3 \right |+5&=7\\ 2\left | x-3 \right |&=2\\ (x-3)&=\pm 1\\ x&=3\pm 1\\ &\begin{cases} x & =4 \\ x & =2 \end{cases}\\ & \end{aligned}&\begin{aligned}-2\left | x-3 \right |+5&=7\\ -2\left | x-3 \right |&=2\\ \left | x-3 \right |&=-1\\ &\\ \textrm{tak}&\: \textrm{ada solusi}\\ &\\ & \end{aligned}\\\hline \end{array}.
\begin{array}{|c|c|c|c|}\hline \textrm{i}&\textrm{j}&\textrm{k}&\textrm{l}\\\hline \begin{aligned}-2\left | -(x-3) \right |+\left | x-3 \right |+5&=7\\ -2\left | x-3 \right |+\left | x-3 \right |&=2\\ -\left | x-3 \right |&=2\\ \left | x-3 \right |&=-2\\ &\\ \textrm{tak ada so}&\textrm{lusi}\\ &\\ &\\ & \end{aligned}&\begin{aligned}\left | x-2 \right |&=\left | x+1 \right |\\ (x-2)&=\pm (x+1)\\ &\begin{cases} (x-2) & =(x+1) \\ &\textrm{tak ada solusi}\\ \textrm{atau}&\\ (x-2)& =-(x+1)\\ \qquad x&=-x-1+2\\ \qquad 2x&=1\\ \qquad x&=\displaystyle \frac{1}{2} \end{cases} \end{aligned}&\begin{aligned}\left | 2-x \right |&=\left | x+1 \right |\\ \left | x-2 \right |&=\left | x+1 \right |\\ &\\ \textrm{selanj}&\textrm{utnya sama}\\ \textrm{denga}&\textrm{n langkah}\\ \textrm{pada}\: \: &\textrm{poin j}\\ &\\ &\\ & \end{aligned}&\begin{aligned}\left | 2x-1 \right |&=\left | x+1 \right |\\ (2x-1)&=\pm (x+1)\\ &\\ \textrm{silahkan}&\textrm{ lanjutkan}\\ \textrm{sendiri}\: \: \: &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}.
C. Pertidaksamaan Nilai Mutlak linear Satu Variabel
\begin{array}{|c|c|c|c|}\hline \multicolumn{4}{|l|}{.}\\ \multicolumn{4}{|c|}{\left | x \right |=\sqrt{x^{2}}}\\ \multicolumn{4}{|r|}{.}\\\hline &&&\\ \left | x \right |< a&\left | x \right |\leq a&\left | x \right |> a&\left | x \right |\geq a\\ &&&\\\hline &&&\\ \Downarrow&\Downarrow&\Downarrow&\Downarrow\\ &&&\\\hline &&&\\ -a< x< a&-a\leq x\leq a&x< -a\quad \textrm{atau}\quad x> a&x\leq -a\quad \textrm{atau}\quad x\geq a\\ &&&\\\hline \end{array}.
\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.
\begin{array}{llll}\\ \multicolumn{4}{l}{\textrm{Tentukanlah himpunan penyelesaian soal berikut}}\\ \textrm{a}.&\left | 2x-3 \right |=7\qquad &\textrm{f}.&\left | 2x-3 \right |=-7\\ \textrm{b}.&\left | 2x-3 \right |< 7&\textrm{g}.&\left | 2x-3 \right |< -7\\ \textrm{c}.&\left | 2x-3 \right |\leq 7&\textrm{h}.&\left | 2x-3 \right |\leq -7\\ \textrm{d}.&\left | 2x-3 \right |> 7&\textrm{i}.&\left | 2x-3 \right |> -7\\ \textrm{e}.&\left | 2x-3 \right |\geq 7&\textrm{j}.&\left | 2x-3 \right |\geq -7 \end{array}.
Jawab:
\begin{array}{|l|l|l|l|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{a}.\quad \left | 2x-3 \right |&=7\\ (2x-3)&=\pm 7\\ 2x&=3\pm 7\\ &\begin{cases} 2x & =3+7 \\ \quad x&=5\\ 2x & =3-7\\ \quad x&=-2 \end{cases}\\ \textbf{HP}&=\left \{ -2,5 \right \}\\ & \end{aligned}}&\multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{f}.\quad \left | 2x-3 \right |&=-7\\ &\\ \textbf{HP}&=\left \{ \: \: \: \right \}\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}}\\\hline \begin{aligned}\textrm{b}.\quad &\left | 2x-3 \right |< 7\\ &-7< 2x-3< 7\\ &-7+3< 2x-3+3< 7+3\\ &-4< 2x< 10\\ &-2< x< 5\\\\ &\textbf{HP}=\left \{ x|-2< x< 5,\: x\in \mathbb{R} \right \}\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{c}.\quad&\textrm{silahkan}\\ &\textrm{selesaikan}\\ &\textrm{sendiri}\\ &\textrm{sebagai}\\ &\textrm{latihan}\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\begin{aligned} \end{aligned}\textrm{d}.\quad &\left | 2x-3 \right |> 7\\ &\begin{cases} 2x-3 & < -7 \\ \quad 2x&< -4\\ \quad x&< -2\\ \textrm{atau}&\\ 2x-3 & > 7\\ \quad 2x&> 10\\ \quad x&> 5 \end{cases}\\ &\\ &\textbf{HP}=\left \{ x|x< -2\: \: \textrm{atau}\: \: x> 5 ,\: x\in \mathbb{R}\right \} \end{aligned}&\begin{aligned}\textrm{g}.\quad &\left | 2x-3 \right |< -7\\ &\\ &\textbf{HP}=\left \{ \: \: \: \right \}\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}

  • Share:

You Might Also Like

0 komentar