Contoh Soal Persamaan dan Pertidaksamaan Nilai Mutlak

\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai untuk}\: \: -\left | -3 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-3\quad &&&\textrm{d}.&-3^{-1}\\ \textrm{b}.&3&\textrm{c}.&3^{-1}\quad &\textrm{e}.&3^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}-\left | -3 \right |&=-\left ( 3 \right )=-3 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Nilai untuk}\: \: \left | -4 \right |-\left | -6^{2}\times 2 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-68\quad &&&\textrm{d}.&68\\ \textrm{b}.&-40&\textrm{c}.&40\quad &\textrm{e}.&76 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\left | -4 \right |-\left | -6^{2}\times 2 \right |&=\left ( 4 \right )-\left | -36\times 2 \right |\\ &=4-\left | -72 \right |\\ &=4-\left ( 72 \right )\\ &=-68 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Jika diberikan}\: \: m\: \: \textrm{adalah bilangan positif dan}\: \: n\: \: \textrm{adalah sebuah bilangan negatif},\\ &\textrm{maka operasi berikut yang mengahsilkan bilangan negatif adalah... .}\\ &\begin{array}{llllll}\\ \textrm{a}. &\left | m\times n \right |\quad &&&\textrm{d}.&m+\left | n \right |\\ \textrm{b}.&m\times \left | n \right |&\textrm{c}.&\left | m \right |\times n\quad &\textrm{e}.&\left | m \right |-n \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Misalkan}\begin{cases} m & = p\\ n &= -q \end{cases}\\ & \begin{array}{|c|c|c|}\hline \textrm{No}&\textrm{Pernyataan}&\textrm{Keterangan}\\\hline \textrm{a}&\left | m\times n \right |=\left | p\times -q \right |=p\times q&\textrm{Positif}\\\hline \textrm{b}&m\times \left | n \right |=p\times \left | -q \right |=p\times q&\textrm{positif}\\\hline \textrm{c}&\left | m \right |\times n=\left | p \right |\times -q=-p\times q&\textbf{negatif}\\\hline \textrm{d}&m+\left | n \right |=p+\left | -q \right |=p+q&\textrm{positif}\\\hline \textrm{e}&\left | m \right |-n=\left | p \right |-(-q)=p+q&\textrm{positif}\\\hline \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Persamaan nilai mutlak berikut yang bernilai benar adalah... .}\\ &\begin{array}{lll}\\ \textrm{a}. &\left | 2^{3}-3^{2} \right |=3^{2}-2^{3}\\ \textrm{b}.&\left | 3^{4}-4^{3} \right |=4^{3}-3^{4}\\ \textrm{c}.&\left | 4^{5}-5^{4} \right |=5^{4}-4^{5}\\ \textrm{d}.&\left | 5^{6}-6^{5} \right |=6^{5}-5^{6}\\ \textrm{e}.&\textrm{pilihan jawaban a, b, c, dan d tidak ada yang benar} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{array}{|c|l|c|}\hline \textrm{No}&\qquad\qquad\qquad\qquad\qquad\qquad \textrm{Pernyataan}&\textrm{Keterangan}\\\hline \textrm{a}&\left | 2^{3}-3^{2} \right |=\left | 8-9 \right |=9-8=3^{2}-2^{3}&\textbf{benar}\\\hline \textrm{b}&\left | 3^{4}-4^{3} \right |=\left | 81-64 \right |=81-64=3^{4}-4^{3}\neq 4^{3}-3^{4}&\textrm{salah}\\\hline \textrm{c}&\left | 4^{5}-5^{4} \right |=\left | 1024-625 \right |=1024-625=4^{5}-5^{4}\neq 5^{4}-4^{5}&\textrm{salah}\\\hline \textrm{d}&\left | 5^{6}-6^{5} \right |=\left | 15625-7776 \right |=15625-7776=5^{6}-6^{5}\neq 6^{5}-5^{6}&\textrm{salah}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Pernyataan berikut yang benar adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left | m \right |+\left | -m \right |=0\\ \textrm{b}.&\displaystyle \frac{\left | -6 \right |+2\left | 3 \right |}{6}=0\\ \textrm{c}.&\left |7^{2}\times 2 \right |-\left | -3^{2} \right |=107\\ \textrm{d}.&\left | x \right |=3,\: \: \textrm{hanya dapat dipenuhi oleh}\: \: -3\: \: \textrm{dan}\: \: 3\\ \textrm{e}.&\textrm{tanda nilai mutlak berlaku hanya untuk bilangan positif} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|l|c|}\hline \textrm{No}&\qquad\qquad\qquad\qquad\qquad \textrm{Pernyataan}&\textrm{Keterangan}\\\hline \textrm{a}&\left | m \right |+\left | -m \right |=m+m=2m\neq 0&\textrm{salah}\\\hline \textrm{b}&\displaystyle \frac{\left | -6 \right |+2\left | 3 \right |}{6}=\displaystyle \frac{6+6}{6}=2\neq 0&\textrm{salah}\\\hline \textrm{c}&\left |7^{2}\times 2 \right |-\left | -3^{2} \right |=98-9=89\neq 107&\textrm{salah}\\\hline \textrm{d}&\left | x \right |=3,\: \: \textrm{hanya dapat dipenuhi oleh}\: \: -3\: \: \textrm{dan}\: \: 3&\textbf{benar}\\\hline \textrm{e}&\textrm{tanda nilai mutlak berlaku hanya untuk bilangan positif}&\textrm{salah}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Nilai dari}\: \: \left | 2\pi -5 \right |-\displaystyle \frac{\pi }{2}=...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3\pi -10}{5}\\ \textrm{b}.&\displaystyle \frac{3\pi +5}{2}\\ \textrm{c}.&\displaystyle \frac{-3\pi +5}{2}\\ \textrm{d}.&\displaystyle \frac{-3\pi +10}{2}\\ \textrm{e}.&\displaystyle \frac{3\pi -10}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\left | 2\pi -5 \right |-\displaystyle \frac{\pi }{2}&=\left (2\pi -5 \right )-\displaystyle \frac{\pi }{2},\qquad \textrm{sebagai catatan bahwa}\: \: \: \pi =3,14...\\ &=\displaystyle \frac{2(2\pi )-2(5)-\pi }{2}\\ &=\displaystyle \frac{3\pi -10}{2} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Bentuk singkat dari}\: \: m-6n\: \: \textrm{atau}\: \: 6n-m\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left | 6n+m \right |\\ \textrm{b}.&\left | n-6m \right |\\ \textrm{c}.&\left | m-6n \right |\\ \textrm{d}.&\left | m-n \right |\\ \textrm{e}.&\left | -6n-m \right | \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left | m-6n \right |=\begin{cases} m-6n & ,\textrm{untuk}\: \: m-6n\geq 0 \\ -(m-6n)=6n-m & ,\textrm{untuk}\: \: m-6n< 0 \end{cases} \end{aligned}\end{array}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Nilai untuk}\: \: x\: \: \textrm{yang memenuhi persamaan}\: \: \left | 2x-5 \right |=11\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{hanya}\: \: 3\\ \textrm{b}.&\textrm{hanya}\: \: 8\\ \textrm{c}.&-3\: \: \textrm{atau}\: \: 8\\ \textrm{d}.&3\: \: \textrm{atau}\: \: -8\\ \textrm{e}.&3\: \: \textrm{atau}\: \: 8 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left | 2x-5 \right |&=11\\ (2x-5)&=\pm 11\\ 2x&=5\pm 11\\ 2x&=\begin{cases} 5+11 & =16 \\ \textrm{atau}&\\ 5-11 &=-6 \end{cases}\\ x&=\begin{cases} 8 & \\ &\textrm{atau} \\ -3 & \end{cases} \end{aligned}\end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Himpunan penyelesaian untuk}\: \: -2\left | x-5 \right |=8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 0 \right \}\\ \textrm{b}.&\left \{ \: \: \right \}\\ \textrm{c}.&\left \{ 4,5 \right \}\\ \textrm{d}.&\left \{ 1,9 \right \}\\ \textrm{e}.&\left \{ -1,9 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}-2\left | x-5 \right |&=8\\ \left | x-5 \right |&=-4\: ,\quad \textrm{karena bernilai negatif maka tidak ada harga}\: \: x\: \: \textrm{yang memenuhi}\\\\ \therefore \: \: \: \textbf{HP}&=\left \{ \: \: \right \} \end{aligned}\end{array}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Nilai}\: \: k\: \: \textrm{yang memenuhi}\: \: \left | 4k \right |=16\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&\pm 2\\ \textrm{d}.&\pm 4\\ \textrm{e}.&\pm 8 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\left | 4k \right |&=16\\ (4k)&=\pm 16\\ k&=\pm \displaystyle \frac{16}{4}\\ &=\pm 4 \end{aligned}\end{array}.

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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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