Contoh Soal Lanjutan 4 Nilai Mutlak

$\begin{array}{l}\\ 21.& \text{Jumlah akar-akar persamaan}\\ & x^2+\left|x\right|-6=0,\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{x}^2+\left|x\right|& -6=0\\ {\left|x\right|}^2+\left|x\right|& -6=0\\ (\left|x\right|+3)& (\left|x\right|-2)=0\\ \left|x\right|=-3& \text{atau}\left|x\right|=2\\ \text{tidak meme}& \text{nuhi (-3) atau}x=\pm 2\\ & \{\begin{array}{ll}{x}_1& =2\\ x_2& =-2\end{array}\\ x_1+x_2& =2+(-2)=0\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Penyelesaian untuk}-2|x-5|=8\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \left\{0\right\}\\ \text{b}.& \left\{\right\}\\ \text{c}.& \{4,5\}\\ \text{d}.& \{1,9\}\\ \text{e}.& \{-1,9\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}-2|x-5|& =8\\ |x-5|& =-4\\ \text{Karena pe}& \text{rsamaan bernilai negatif},\\ \text{maka tida}& \text{k ada nilai}x\text{yang memenuhi}\\ \therefore \mathbf{\text{HP}}& =\left\{\right\}\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Nilai}x\text{yang memenuhi}\left|{\displaystyle \frac{x-5}{2x+1}}\right|=2\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{5}{3}\text{atau}\frac{3}{5}}\\ \text{b}.& {\displaystyle \frac{5}{3}\text{atau}-\frac{3}{5}}\\ \text{c}.& -{\displaystyle \frac{7}{3}\text{atau}\frac{3}{5}}\\ \text{d}.& {\displaystyle \frac{7}{3}\text{atau}-\frac{3}{5}}\\ \text{e}.& -{\displaystyle \frac{7}{3}\text{atau}\frac{3}{7}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{x-5}{2x+1}}\right|& =2\\ {\displaystyle \frac{x-5}{2x+1}}& =\pm 2\\ \text{untuk}& x=2\\ {\displaystyle \frac{x-5}{2x+1}}& =2\\ x-5& =2(2x+1)\\ x-4x& =2+5\\ -3x& =7\\ x& ={\displaystyle \frac{7}{-3}=-\frac{7}{3}}\\ \text{untuk}& x=-2\\ x-5& =-2(2x+1)\\ x+4x& =-2+5\\ 5x& =3\\ x& ={\displaystyle \frac{3}{5}}\end{array}\end{array}$

$\begin{array}{l}\\ 24.& \text{Nilai}x\text{yang memenuhi}||5x-4|-3|=2\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -3\text{atau}-6\\ \text{b}.& -3\text{atau}6\\ \text{c}.& 3\text{atau}-6\\ \text{d}.& 3\text{atau}6\\ \text{e}.& 6\text{atau}9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}||2x-3|-4|& =5\\ |2x-3|-4& =\pm 5\\ |2x-3|& =\pm 5+4\text{maka},\\ |2x-3|& =5+4=9(\text{memenuhi})\\ |2x-3|& =-5+4=-1\\ & (\mathbf{\text{tidak memenuhi}})\\ \text{Selanjutnya}& ,\\ (2x-3)& =\pm 9\\ 2x& =\pm 9+3\\ \text{untuk}& x=9\\ 2x& =9+3\\ x& ={\displaystyle \frac{12}{2}=6}\\ \text{untuk}& x=-9\\ 2x& =-9+3\\ x& ={\displaystyle \frac{-6}{2}=-3}\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Nilai}x\text{yang memenuhi}\\ & \text{persamaan}|x+1|-2|x-3|=5|x-4|\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1\frac{1}{6}\text{atau}1\frac{2}{3}}\\ \\ \text{b}.& {\displaystyle 1\frac{5}{6}\text{atau}2\frac{2}{5}}\\ \\ \text{c}.& {\displaystyle 1\frac{5}{6}\text{atau}2\frac{2}{3}}\\ \\ \text{d}.& {\displaystyle 2\frac{3}{4}\text{atau}4\frac{2}{3}}\\ \\ \text{e}.& {\displaystyle 3\frac{1}{4}\text{atau}4\frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\text{e}\\ & \begin{array}{rl}|x+1|-& 2|x-3|=5|x-4|\\ \text{Perhatik}& \text{an untuk batas sesuai definisi, maka}\\ x=-1& ,x=3,x=4\\ \text{saat}x& \ge 4\\ |x+1|& =x+1,|x-3|=x-3,|x-4|=x-4\\ (x+1)-& 2(x-3)=5(x-4)\\ x-2x-& 5x=-1-6-20\\ -6x& =-27\\ x& ={\displaystyle \frac{27}{6}=4\frac{3}{6}(\text{memenuhi})}\\ \text{saat}3& \le x<4\\ |x+1|& =x+1,|x-3|=x-3,|x-4|=4-x\\ (x+1)-& 2(x-3)=5(4-x)\\ x-2x+& 5x=-1-6+20\\ 4x& =13\\ x& ={\displaystyle \frac{13}{4}=3\frac{1}{4}(\text{memenuhi})}\\ \text{saat}-& 1\le x<3\\ |x+1|& =x+1,|x-3|=3-x,|x-4|=4-x\\ (x+1)-& 2(3-x)=5(4-x)\\ x+2x+& 5x=-1+6+20\\ 8x& =25\\ x& ={\displaystyle \frac{25}{8}=3\frac{1}{8}(\text{tidak memenuhi})}\\ \text{saat}-& 1>x\\ |x+1|& =-x-1,|x-3|=3-x,|x-4|=4-x\\ (-x-1)& -2(3-x)=5(4-x)\\ -x+2x& +5x=1+6+20\\ 6x& =27\\ x& ={\displaystyle \frac{27}{6}=4\frac{3}{6}(\text{tidak memenuhi})}\end{array}\end{array}$

Sumber Referensi

1. Kanginan, M. 2016. Matematika untuk Siswa SMA-MA/SMK-MAK Kelas X. Bandung: Srikandi Empat Widya Utama.
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.