A BAB 2 Persamaan dan Pertidaksamaan Nilai Mutlak Linear Satu Variabel - Media Pembelajaran Namal Sparkly Santa Hat Ice Cream

Persamaan dan Pertidaksamaan Nilai Mutlak Linear Satu Variabel BAB I Persamaan dan Pertidaksamaan Nilai Mutlak Linear Satu ...

BAB 2 Persamaan dan Pertidaksamaan Nilai Mutlak Linear Satu Variabel

BAB 2 Persamaan dan Pertidaksamaan Nilai Mutlak Linear Satu Variabel

BAB 2 Persamaan dan Pertidaksamaan Nilai Mutlak Linear Satu Variabel

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Persamaan dan Pertidaksamaan Nilai Mutlak Linear Satu Variabel


BAB I Persamaan dan Pertidaksamaan Nilai Mutlak Linear Satu Variabel (Kelas X K13 Revisi)

Posted on Agustus 5, 2017by 
Sebelumnya Anda bisa baca di sini
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}

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