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

Sebelumnya Anda bisa baca di sini

A. Nilai Mutlak

Nilai mutlak suatu bilangan adalah adalah jarak suatu bilangan yang diukur dari posisi O dari 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}.

Soal yang belum diselesaikan silahkan dicoba dikerjakan sebagai latihan mandiri.

Sumber Referensi

  1. Kementerian Pendidikan dan Kebudayaan Republik Indonesia. 2016. Matematika Kelas X SMA/MA/SMK/MAK Edisi Revisi 2016. Jakarta: Kementerian Pendidikan dan Kebudayaan.
  2. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, 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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