Lanjutan 3 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Wajib

Dalam menentukan batas interval pada contoh soal-soal penyelesaian berikut bisa merujuk di sini dan di sini (untuk yang berkaitan dengan nilai mutlak), moga dapat membantu Anda dalam memahami interval dalam tulisan berikut

\begin{array}{ll}\\ \fbox{22}.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \left | \displaystyle \frac{5}{4x-3} \right |\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{A}.&-\displaystyle \frac{1}{2}\leq x< \frac{3}{4}\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{B}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \frac{3}{4}< x\leq 2\\ \textrm{C}.&-\displaystyle \frac{1}{2}\leq x\leq 2,\: x\neq \frac{3}{4}\\ \textrm{D}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x>\frac{3}{4}\\ \textrm{E}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x\geq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{E}\\ &\begin{aligned}\left | \displaystyle \frac{5}{4x-3} \right |&\leq 1\\ 5&\leq \left | 4x-3 \right |\\ \left | 4x-3 \right |&\geq 5\\ \left | 4x-3 \right |^{2}&\geq 5^{2}\\ (4x-3)^{2}-5^{2}&\geq 0\\ (4x-3+5)(4x-3-5)&\geq 0\\ (4x+2)(4x-8)&\geq 0\\ 8(2x+1)(x-2)&\geq 0\\ (2x+1)(x-2)&\geq 0\\ \therefore \: x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x&\geq 2 \end{aligned} \end{array}.
\begin{array}{ll}\\ \fbox{23}.&\textrm{(SPMB2002)Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: x\left | x \right |-2< x-2\left | x \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{A}.&x<-2\: \: \textrm{atau}\: \: -1<x<1\\ \textrm{B}.&-2<x<-1\: \: \textrm{atau}\: \: x>2\\ \textrm{C}.&-2<x<1\: \: \textrm{atau}\: \: x>2\\ \textrm{D}.&-1<x<1\: \: \textrm{atau}\: \: x>2\\ \textrm{E}.&-1<x<2\: \: \textrm{atau}\: \: x>3 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}x\left | x \right |-2< x-2\left | x \right |&\\ x\left | x \right |+2\left | x \right |-x-2&< 0\\ (x+2)\left | x \right |-(x+2)&<0\\ (x+2)\left ( \left | x \right |-1 \right )&<0,\: \: \textrm{atau}\: \: \left ( \left | x \right |-1 \right )(x+2)<0\\ \textrm{Kemungkinan pertama}&\: \textrm{yaitu}:\\ (x+2)>0\: \: \textrm{atau}\: \: \left | x \right |-1&<0\\ x>-2\: \: \textrm{atau}\: \: \left | x \right |&<1\Rightarrow -1<x<1\\ \textrm{Kemungkinan yang ke}&\textrm{dua},\\ \left | x \right |-1>0\: \: \textrm{atau}\: \: (x+2)&<0\\ \left | x \right |>1\: \: \textrm{atau}\: \: x<&-2\: \Rightarrow x<-2\\ \therefore ,\: x<-2\: \: \textrm{atau}\: \: -1<x&<1 \end{aligned} \end{array}

Lanjutan 2 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Wajib

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\begin{array}{ll}\\ \fbox{17}.&\textrm{Pernyataan berikut yang tepat adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{b}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<1\\ \textrm{c}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: -3<m<-2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{d}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: 2<m<3\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{e}.&\textrm{pilihan jawaban baik a, b, c, maupun d tidak ada yang benar} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikanlah opsi}\: \: \textbf{a}\: ,\\ \left | m \right |&<2\\ -2<m&<2\\ \textrm{sehing}&\textrm{ga untuk nilai}\: \: m\in \mathbb{R}\: \: \textrm{pada rentang}\\ -1<m&<2\: \: \: \: \textrm{akan memenuhi semua}\\\\ \textbf{silahk}&\textbf{an cek sendiri untuk opsi jawaban yang lain} \end{aligned} \end{array}
\begin{array}{ll}\\ \fbox{18}.&\textrm{Himpunan penyelesaian dari pertidaksamaan}\: \: \left | x+3 \right |<2\left | x-4 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&\left \{ x|x<-\displaystyle \frac{5}{3} \right \}&&&\textrm{d}.&\left \{ x|x<\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x>11 \right \}\\\\ \textrm{b}.&\left \{ x|\: \displaystyle \frac{5}{3}<x<-11 \right \} \quad&\textrm{c}.&\left \{ x|x\geq -11 \right \} \quad&\textrm{e}.&\left \{ x|x>-\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x<-11 \right \} \\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\left | x+3 \right |&<2\left | x-4 \right |\\ \left ( x+3 \right )^{2}&<2^{2}\left ( x-4 \right )^{2}\quad \textrm{dikuadratkan masing-masing ruas}\\ x^{2}+6x+9&<4\left ( x^{2}-8x+16 \right )\\ x^{2}-4x^{2}+6x+32x+9-64&<0\\ -3x^{2}+38x-55&<0\\ 3x^{2}-38x+55&>0\\ \left ( 3x-5 \right )\left (x -11 \right )&>0\\\\ \textrm{Berikut untuk}&\: \textrm{garis bilangannya}\\ &\begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{1-3}\cline{8-9} +&+&&&&&&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{r}{\begin{matrix} \displaystyle \frac{3}{5}\\ \end{matrix}}&&&\multicolumn{2}{l}{11}&\\ \end{array} \end{aligned} \end{array}
\begin{array}{ll}\\ \fbox{19}.&\textrm{Himpunan penyelesaian dari pertidaksamaan}\: \: \left | x^{2}+5x \right |\leq 6\: \: \textrm{adalah... .}\\ &\textrm{a}.\quad \left \{ x|-6\leq x\leq 1 \right \}\\ &\textrm{b}.\quad \left \{ x|-3\leq x\leq -2 \right \}\\ &\textrm{c}.\quad \left \{ x|-6\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 1\right \}\\ &\textrm{d}.\quad \left \{ x|-6\leq x\leq -5 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\ &\textrm{e}.\quad \left \{ x|-5\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \left | x^{2}+5x \right |&\leq 6\\ -6\leq x^{2}+5x&\leq 6\\ & \end{aligned}}\\\hline -6\leq x^{2}+5x&x^{2}+5x\leq 6\\\hline \begin{aligned}x^{2}+5x+6&\geq 0\\ (x+3)(x+2)&\geq 0 \end{aligned}&\begin{aligned}x^{2}+5x-6&\leq 0\\ (x+6)(x-1)&\leq 0 \end{aligned}\\\hline &\\ \begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{1-3}\cline{8-9} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-3}&&&\multicolumn{2}{r}{-2}&\\ \end{array}&\begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{4-7} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-6}&&&\multicolumn{2}{c}{1}&\\ \end{array} \\ &\\\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Kesimpulan}:&\\ &\begin{array}{ccc|cccc|ccc|ccc|cccccccc}\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{2}{c}{.}&&\multicolumn{2}{l}{.}&&\multicolumn{2}{r}{.}\\\cline{4-7}\cline{11-13} &&&&&&&&&&&&&&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-6}&&&\multicolumn{2}{r}{-3}&&\multicolumn{2}{l}{-2}&&\multicolumn{2}{l}{1}&\\ \end{array} \\ & \end{aligned}}\\\hline \end{array} \end{array}
\begin{array}{ll}\\ \fbox{20}.&\textbf{(USM UGM Mat IPA)}\textrm{Semua nilai \textit{x} yang memenuhi}\: \: x\left | x-2 \right |<x-2\: \: \textrm{adalah... .}\\ &\textrm{a}.\quad x<-1\: \: \textrm{atau}\: \: 1<x<2\\ &\textrm{b}.\quad x<-2\\ &\textrm{c}.\quad -2<x<-1\\ &\textrm{d}.\quad x<-1\\ &\textrm{e}.\quad -2<x<1\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ &x\left | x-2 \right |<x-2\\ & \end{aligned}}\\\hline x<2&x\geq 2\\\hline \begin{aligned}x\left | x-2 \right |&<x-2\\ x(2-x)&<x-2\\ 2x-x^{2}&<x-2\\ -x^{2}+x+2&<0\\ x^{2}-x-2&>0\\ (x-2)(x+1)&>0\end{aligned}&\begin{aligned}x\left | x-2 \right |&<x-2\\ x(x-2)&<x-2\\ x^{2}-2x&<x-2\\ x^{2}-3x+2&<0\\ (x-1)(x-2)&<0\\ & \end{aligned}\\ \begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{1-3}\cline{8-9} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-1}&&&\multicolumn{2}{c}{2}&\\ \end{array}&\begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{4-7} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{c}{1}&&&\multicolumn{2}{c}{2}&\\ \end{array} \\\hline \textbf{ada yang memenuhi}&\textbf{tidak ada yang memenuhi}\\\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{yang memenuhi}:&\\ &x<-1\\ & \end{aligned}}\\\hline \end{array} \end{array}

Sumber Referensi

  1. Budhi, W. S. 2014. Bupena Matematika SMA/MA Kelas X Kelompok Wajib. Jakarta: Erlangga.
  2. Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: Grasindo.
  3. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.

Lanjutan 1 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Wajib

\color{magenta}\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}\\ &\color{black}\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{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}.

\begin{array}{ll}\\ \fbox{11}.&\textrm{Nilai}\: \: p\: \: \textrm{yang memenuhi}\: \: 10-4\left | 4-5p \right |=26\: \: \textrm{adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&2\: \: \textrm{atau}\: \: 1\displaystyle \frac{2}{3}&&&\textrm{d}.&1\: \: \textrm{atau}\: \: -\displaystyle \frac{3}{5}\\\\ \textrm{b}.&2\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\quad&\textrm{c}.&-2\displaystyle \frac{3}{4}\: \: \textrm{atau}\: \: 1\quad&\textrm{e}.&-1\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}10-4\left | 4-5p \right |&=-26\\ -4\left | 4-5p \right |&=-36\\ \left | 4-5p \right |&=9\\ (4-5p)&=\pm 9\\ -5p&=-4\pm 9\\ p&=\displaystyle \frac{-4\pm 9}{-5}\\ p&=\begin{cases} \displaystyle \frac{-4+9}{-5} & =-1 \\ \textrm{atau} & \\ \displaystyle \frac{-4-9}{-5} & = \displaystyle \frac{13}{5}=2\frac{3}{5} \end{cases} \end{aligned} \end{array}.

\begin{array}{ll}\\ .\: \: \: \: \: \: &\textrm{Persamaan yang memenuhi rumus tersebut adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=\left | -x-2 \right |\\ \textrm{b}.&y=-2x-4\\ \textrm{c}.&y=-\left | 2x-4 \right |\\ \textrm{d}.&y=\left | -2x-4 \right |\\ \textrm{e}.&y=\left | -2x+4 \right | \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Dengan cara substitusi langsung kita akan mendapatkan}\\ &\bullet \quad \textrm{untuk}\: \: x=4\: \: \textrm{menyebabkan nilai}\: \: y=4\: \: \textrm{dan}\\ &\qquad \textrm{sampai langkah di sini hanya ada 1 persamaan yang memenuhi yaitu}:\: \: y=\left | -2x+4 \right | \end{aligned}\end{array}.

\begin{array}{lll}\\ \fbox{13}.&\textrm{Gambarlah garfik untuk persamaan}\: \: \left | x \right |+\left | y \right |=4\\\\\\ &\textrm{Jawab}:\\\\ &\begin{array}{|cc|cc|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\left | x \right |+\left | y \right |=4\\ & \end{aligned}}\\\hline x> 0\: ,\: y> 0&&&x> 0\: ,\: y<0\\\hline x+y=4&&&x+(-y)=4\\ &&&\\\hline &&&\\ x<0\: ,\: y> 0&&&x<0\: ,\: y<0\\\hline (-x)+y=4&&&(-x)+(-y)=4\\\hline \end{array} \end{array}

Berikut sebagai ilustrasinya

\begin{array}{lll}\\ \fbox{14}.&\textrm{Tentukanlah nilai}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\: \: x+\left | x \right |+y=5\: \: \textrm{dan}\: \: x+\left | y \right |-y=10\\\\ &\textrm{Jawab}:\\\\ &\begin{array}{|cc|cc|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\left | x \right |+x+y=5\\ &\: \: \qquad \textrm{dan}\\ &x+\left | y \right |-y=10\\ & \end{aligned}}\\\hline x> 0\: ,\: y> 0\quad \textbf{(kuadran I)}&&&x> 0\: ,\: y<0\quad \textbf{(kuadran IV)}\\\hline \begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5 \\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\: (\textrm{ada}) \end{cases}&&&\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5\\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases}\\ &&&\\\hline &&&\\ x<0\: ,\: y> 0\quad \textbf{(kuadran II)}&&&x<0\: ,\: y<0\quad \textbf{(kuadran III)}\\\hline \begin{cases} \left | x+x+y=5 \right | & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\: \textbf{(tidak memenuhi)} \end{cases}&&&\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\: \textbf{(tidak memenuhi)}\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases}\\\hline \end{array} \end{array}

. Beriut ilustrasi grafiknya

Jadi, nilai x dan y yang memenuhi untuk persamaan di atas adalah  x = 4  dan  y = -3

\color{magenta}\begin{array}{ll}\\ \fbox{15}.&\textrm{Gambarlah grafik fungsi mutlak dari}\\ &\textrm{a}.\quad y=\left | x-2 \right |\\ &\textrm{b}.\quad y=-\left | x-2 \right |\\ &\textrm{c}.\quad y=2+\left | x-2 \right |\\ &\textrm{d}.\quad y=2-\left | x-2 \right |\\ &\textrm{e}.\quad y=\left | 2+\left | x-2 \right | \right |\\ &\textrm{f}.\quad y=\left | 2-\left | x-2 \right | \right | \end{array}.

Berikut grafik dari soal di atas

untuk Soal poin a)
untuk jawaban poin b)
untuk jawaban poin c)
untuk jawaban poin d)
untuk jawaban poin f)
sSedangkan poin e) silahkan tentukan atau buat sendiri grafiknya

Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Wajib

\color{magenta}\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai untuk}\: \: -\left | -2019 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-2019\quad &&&\textrm{d}.&-2019^{-1}\\ \textrm{b}.&2019&\textrm{c}.&2019^{-1}\quad &\textrm{e}.&2019^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\color{blue}\begin{aligned}-\left | -2019 \right |&=-\left ( 2019 \right )=-2019 \end{aligned} \end{array}.

\color{magenta}\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}\\ &\color{black}\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}.

\color{magenta}\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... .}\\ &\color{black}\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}\\ &\color{blue}\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}.

\color{magenta}\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}\\ &\color{balck}\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}.

\color{magenta}\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}\\ &\color{black}\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}.

Lanjutan 2 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Peminatan

\begin{array}{ll}\\ \fbox{21}.&\textrm{Solusi dari persamaan eksponen}\\ & \left ( x^{2}-5x+5 \right )^{2x+3}=\left ( x^{2}-5x+5 \right )^{3x-2}\: \: \textrm{adalah}\: ....\\\\ &\begin{matrix} \textrm{A}. & 1,2,4,5, \displaystyle \frac{9-\sqrt{5}}{2},\frac{9+\sqrt{5}}{2}\\ & & \\ \textrm{B}. & 1,3,4,5, \displaystyle \frac{9-\sqrt{5}}{2},\frac{9+\sqrt{5}}{2} \\ & & \\ \textrm{C}. & 1,2,4,5, \displaystyle \frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2} \\ & & \\ \textrm{D}. & 1,3,4,5, \displaystyle \frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2} \\ & & \\ \textrm{E}. & 1,2,3,6, \displaystyle \frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2} \\ & & \end{matrix}\\\\\\ &\textrm{Jawab}\: \: \textbf{D}\\ \end{array}.

.\quad\: \, \color{blue}\begin{aligned}\left (x^{2}-5x+5 \right )^{\left ( 2x+3 \right )}&=\left (x^{2}-5x+5 \right )^{\left ( 3x-2 \right )}\\ \textbf{bentuk umum}:\: &\: h(x)^{f(x)}=h(x)^{g(x)}\\ \bullet \: \: \textrm{Pangkat sama}&, \\ 2x+3&=3x-2\\ 2x-3x&=-2-3\\ -x&=-5\Rightarrow x=5\\ \bullet \: \: \textrm{Basis}\: h(x)=1&\\ x^{2}-5x+5&=1\\ x^{2}-5x+4&=0\\ (x-1)(x-4)&=0\Rightarrow x_{1}=1\: \: \textrm{atau}\: \: x_{2}=4\\ \bullet \: \: \textrm{Basis}\: h(x)=0&\: \textrm{dengan syarat},f(x)\&g(x)> 0\\ f(1)&=2.1+3>0\\ g(1)&=3.1-2>0,\: \: dan\\ f(4)&=2.4+3>0\\ g(4)&=3.4-2>0\\ \textrm{Sehingga keduanya}&\: \textrm{memenuhi} \end{aligned}.

.\quad\: \, \color{red}\begin{aligned} \bullet \: \: \textrm{Basis}\: h(x)=0&\: \textrm{dengan syarat},f(x)\&g(x)> 0\\ x^{2}-5x+5&=0\Rightarrow x_{1,2}=\displaystyle \frac{-(-5)\pm \sqrt{5^{2}-4.1.5}}{2}\\ &\Rightarrow x_{1,2}=\displaystyle \frac{5\pm \sqrt{5}}{2}\\ f\left ( \displaystyle \frac{5+\sqrt{5}}{2} \right )&=2\left ( \displaystyle \frac{5+\sqrt{5}}{2} \right )+3>0\\ g\left ( \displaystyle \frac{5-\sqrt{5}}{2} \right )&=3\left ( \displaystyle \frac{5-\sqrt{5}}{2} \right )-2>0 \end{aligned}.

.\quad\: \, \color{magenta}\begin{aligned} \bullet \: \: \textrm{Basis}\: h(x)=-1&\: \textrm{dengan syarat},f(x)\&g(x)\\ &\textrm{keduanya ganjil atau keduanya genap}\\ x^{2}-5x+5&=-1\\ x^{2}-5x+6&=0\\ (x-2)(x-3)&=0\Leftrightarrow x_{1}=2\: \: \textrm{atau}\: \: x_{2}=3\\ \textrm{untuk}\: \: \: f(2)&=2.2+3=7\Rightarrow \textrm{ganjil}\\ g(2)&=3.2-2=6-2=4\Rightarrow \textrm{genap}\\ \textrm{sehingga}&\: \: x=2\: \: \textrm{bukan penyelesaian}\\ \textrm{untuk}\: \: \: f(3)&=2.3+3=9\Rightarrow \textrm{ganjil}\\ g(3)&=3.3-2=9-2=7\Rightarrow \textrm{ganjil}\\ maka\: \: x&=3\: \: \textrm{merupakan penyelesaian}.\\ \textrm{Jadi, solusi dari}&\: \textrm{persamaan di atas adalah}\\ &=1,3,4,5,\displaystyle \frac{5\pm \sqrt{5}}{2} \end{aligned}.

\color{blue}\begin{array}{ll}\\ \fbox{22}.&\textrm{Jumlah akar-akar persamaan}\: \: 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{A}.&-2&&&\textrm{D}.&1\\ \textrm{B}.&-1&\textrm{C}.&0&\textrm{E}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=1 \end{aligned} \end{array}.

\color{magenta}\begin{array}{ll}\\ \fbox{23}.&\textrm{(\textbf{Prestasi Maksima 2007}) Carilah semua bilangan real}\: \: x \\ &\textrm{yang memenuhi persamaan eksponensial}\: \: \: \: \displaystyle \frac{8^{x}+27^{x}}{12^{x}+18^{x}}=\frac{7}{6}\\\\ &\textrm{Solusi}:\\ &\color{black}\begin{aligned}\textrm{Untuk jawaban No}.23\quad\qquad\quad & \\\\ \displaystyle \frac{8^{x}+27^{x}}{12^{x}+18^{x}}&=\frac{7}{6}\\ \displaystyle \frac{2^{3x}+3^{3x}}{(2^{2}.3)^{x}+(2.3^{2})^{x}}&=\frac{7}{6}\Leftrightarrow \displaystyle \frac{2^{3x}+3^{3x}}{2^{2x}.3^{x}+2^{x}.3^{2x}}=\frac{7}{6}\\ \displaystyle \frac{\left (2^{x} \right )^{3}+\left (3^{x} \right )^{3}}{\left (2^{x} \right )^{2}.3^{x}+2^{x}.\left (3^{x} \right )^{2}}&=\frac{7}{6}\\ \begin{cases} 2^{x} &=p \\ 3^{x} &=q \end{cases}\\ \displaystyle \frac{p^{3}+q^{3}}{p^{2}q+pq^{2}}&=\frac{7}{6}\Leftrightarrow \displaystyle \frac{(p+q)\left ( p^{2}-pq+q^{2} \right )}{pq(p+q)}=\frac{7}{6}\\ \displaystyle \frac{p^{2}-pq+q^{2}}{pq}&=\frac{7}{6}\\ 6p^{2}-6pq+6q^{2}&=7pq\\ 6p^{2}-13pq+6q^{2}&=0\\ (2p-3q)(3p-2q)&=0\\ 2p-3q=0\quad \textrm{V}\quad 3p-2q&=0\\ 2p=3q\quad \textrm{V}\quad 3p&=2q\\ 2.2^{x}=3.3^{x}\quad \textrm{V}\quad 3.2^{x}&=2.3^{x}\\ x=-1\qquad \textrm{V}\qquad x&=1\\ \textrm{Jadi},\: \: \textrm{HP}=\left \{ x|x=-1\: \: \textrm{atau}\: \: x=1 \right \}& \end{aligned} \end{array}.

\color{magenta}\begin{array}{ll}\\ \fbox{24}.&\textrm{Jika}\: \: f(x)=b^{x},\: \: \textrm{di mana konstan positif},\\\\ &\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}= ....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&f\left ( x^{2} \right )&&&\textrm{D}.&f(x+1)-f(x-1)\\ \textrm{B}.&f(x+1)f(x-1)&\textrm{C}.&f(x+1)+f(x-1)&\textrm{E}.&f\left ( x^{2}-1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{E}\\ &\color{black}\begin{aligned}\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}&=\frac{b^{x^{2}+x}}{b^{x+1}}\\ &=b^{x^{2}+x-(x+1)}\\ &=b^{x^{2}-1}\\ &=f\left ( x^{2}-1 \right ) \end{aligned} \end{array}..

\color{magenta}\begin{array}{ll}\\ \fbox{25}.&\textrm{Misal}\: \: n\: \: \textrm{suatu bilangan asli dan}\: \: x\: \: \textrm{adalah bilangan riil positif}\\ &\textrm{Jika}\: \: \: \: 2x^{n}+\displaystyle \frac{3}{x^{.^{-\displaystyle \frac{n}{2}}}}-2=0,\: \textrm{maka nilai}\: \: \: \displaystyle \frac{2}{x^{n}+\displaystyle \frac{1}{4}}=....\\ &\textrm{Solusi}:\\ &\color{black}\begin{aligned}2x^{n}+\displaystyle \frac{3}{x^{.^{-\displaystyle \frac{n}{2}}}}-2&=0\\ 2\left ( x^{.^{\displaystyle \frac{n}{2}}} \right )^{2}+3.\left ( x^{.^{\displaystyle \frac{n}{2}}} \right )-2&=0\\ \textrm{Misalkan}\: \: \: \: p=x^{.^{\displaystyle \frac{n}{2}}}&\: \: \textrm{maka}\\ 2p^{2}+3p-2&=0\\ (p+2)(2p-1)&=0\Leftrightarrow p_{1}=-2\: \: \textbf{atau}\: \: p_{2}=\frac{1}{2}\\ \textrm{Ambil nilai}\: \: p\: \: \textrm{yang}&>0,\: \: \textrm{yaitu}\: \: p=\frac{1}{2}\\ \textrm{maka nilai}\: \: \displaystyle \frac{2}{x^{n}+\displaystyle \frac{1}{4}}&=\displaystyle \frac{2}{p^{2}+\displaystyle \frac{1}{4}}=\displaystyle \frac{2}{\displaystyle \frac{1}{4}+\frac{1}{4}}=\displaystyle \frac{2}{\displaystyle \frac{1}{2}}=4 \end{aligned} \end{array}..

\color{blue}\begin{array}{ll}\\ \fbox{26}.&\textrm{Diketahui persamaan}\\\\ &\displaystyle \frac{x^{2}}{10000}=\displaystyle \frac{1000}{x^{2\left ( {{^{10}\log x}} \right )-8}} \\\\ &\textrm{Hasil kali dari nilai-nilai}\: \: x\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&10^{2}&&&\textrm{D}.&10^{7}\\ \textrm{B}.&10^{3}&\textrm{C}.&10^{4}&\textrm{E}.&10^{8} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}\displaystyle \frac{x^{2}}{10000}&=\displaystyle \frac{1000}{x^{2\left ( ^{10}\log x \right )-8}}\\ x^{2}.x^{2\left ( ^{10}\log x \right )-8}&=10^{3}.10^{4}\\ x^{2\left ( ^{10}\log x \right )-6}&=10^{7}\\ x^{2\log x-6}&=10^{7},\quad \textrm{di}-log-\textrm{kan masing-masing ruas}\\ \log x^{\log x^{2}-6}&= \log 10^{7}\\ \left ( 2\log x-6 \right )\log x&=7\\ 2\left ( \log x \right )^{2}-6\left ( \log x \right )-7&=0\\ \textrm{persamaan kuadrat yan}&\textrm{g variabelnya berupa}-log.\\ \textrm{Sehingga hasil kali dari}\, &\: \textrm{nilai-nilai}\: \: x-\textrm{nya adalah}:\\ \alpha +\beta &=-\displaystyle \frac{b}{a}\\ \log x_{1}+\log x_{2}&=-\displaystyle \frac{b}{a},\qquad \textrm{ingat bahwa}\: \: \log x=\, ^{10}\log x\\ \log x_{1}.x_{2}&=-\displaystyle \frac{-6}{2}\\ ^{10}\log x_{1}.x_{2}&=3\\ x_{1}.x_{2}&=10^{3} \end{aligned} \end{array}.

Sumber Referensi

  1. Kanginan, Marthen, Ghany Akhmad, Hadi Nurdiansyah. 2014. Matematika untuk SMA/MA Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: Yrama Widya.
  2. Soetiyono, dkk. 2007. Matematika Interaktif 3B Sekolah Menengah Atas Kelas XII Program IPA. Jakarta: Yudistira.
  3. Sukino. 2013. Matematika Kelompok Peminatan Matematika dan Ilmu Alam untuk SMA/MA Kelas x. Jakarta: Erlangga.
  4. Susianto, Bambang. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: Grasindo.
  5. Tung, Khoe Yao. 2012. Pintar Matematika SMA Kelas XII IPA Untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

Lanjutan 1 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Peminatan

\color{blue}\begin{array}{ll}\\ \fbox{11}.&\textrm{Jika}\: \: f(x)=x^{3}-3x^{2}-3x-1,\\ &\textrm{maka nilai dari}\: \: f\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&-\sqrt{2}&&&\textrm{D}.&1\\ \textrm{B}.&-1&\textrm{C}.&0&\textrm{E}.&\sqrt{2} \end{array}\\\\ &\qquad\qquad\qquad\qquad\qquad\quad(\textbf{\color{black}\textit{UM UNDIP 2012 Math IPA}})\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}f(x)&=x^{3}-3x^{2}-3x-1\\ &=\left ( x- 1\right )^{3}-6x\\ f\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )&=\left (1+\sqrt[3]{2}+\sqrt[3]{4}-1 \right )^{3}-6\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )\\ &=\left ( \sqrt[3]{2}+\sqrt[3]{4} \right )^{3}-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=\left (\sqrt[3]{2} \left ( \sqrt[3]{2}+1 \right ) \right )^{3}-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=2\left ( 1+3\sqrt[3]{2}+3\sqrt[3]{4}+2 \right )-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=\left ( 6+6\sqrt[3]{2}+6\sqrt[3]{4} \right )-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=0 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{12}.&\textrm{Jika}\: \: \textbf{a}\: \: \textrm{dan}\: \: \textbf{b}\: \: \textrm{adalah bilangan bulat positif yang memenuhi persamaan}\\ &\textbf{a}^{\textbf{b}}=2^{20}-2^{19},\: \textrm{maka nilai dari}\: \: \textbf{a}+\textbf{b}=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&3&&&\textrm{D}.&21\\ \textrm{B}.&7&\textrm{C}.&19&\textrm{E}.&23 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\color{red}\begin{aligned}\textbf{a}^{\textbf{b}}&=2^{20}-2^{19}\\ &=2^{^{^{(19+1)}}}-2^{19}\\ &=2^{19}.2^{1}-2^{19}\\ &=2^{19}(2-1)\\ &=2^{19}\\ \textrm{Se}&\textrm{hingga nilai}\: \: \textbf{a}+\textbf{b}=2+19=21\end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Perhatikan gambar berikut} \end{array}

\color{blue}\begin{array}{ll}\\ .\quad\: \, &\textrm{Persamaan grafik fungsi seperti gambar di atas adalah}\, ....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&f(x)=2^{x-2}&&&\textrm{D}.&f(x)=\, ^{2}\log (x-1)\\ \textrm{B}.&f(x)=2^{x}-2&\textrm{C}.&f(x)=2^{x}-1&\textrm{E}.&f(x)=\, ^{2}\log (x+1) \end{array}\\\\ &\color{black}\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}\textrm{A}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1-2}=2^{-1}=\frac{1}{2}\neq 0\: \: (\textrm{salah})\\ \textrm{B}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1}-2=2^{1}-2=0= 0\: \: (\color{red}\textbf{benar})\\ \textrm{C}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1}-1=2^{1}-1=1\neq 0\: \: (\textrm{salah})\\ \textrm{D}&\Rightarrow (1,0)\Rightarrow f(1)=\, ^{2}\log (1-1)=\, ^{2}\log 0=\textbf{tidak mungkin}\neq 0\: \: (\textrm{salah})\\ \textrm{E}&\Rightarrow (1,0)\Rightarrow f(1)=\, ^{2}\log (1+1)=\, ^{2}\log 2=1\neq 0\: \: (\textrm{salah})\\ \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{15}.&\textrm{Solusi untuk persamaan}\: \: 3^{2x+1}=81^{x-2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&0&&&\textrm{D}.&4\displaystyle \frac{1}{2}\\\\ \textrm{B}.&2&\textrm{C}.&4&\textrm{E}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\begin{array}{|c|c|}\hline \color{red}\begin{aligned}3^{2x+1}&=81^{x-2}\\ \left (3^{2x} \right ).3^{1}&=\left (3^{4} \right )^{x-2}\\ 3.\left (3^{2x} \right )&=3^{4x}.3^{-8}\\ 3.\left (3^{2x} \right )&=\displaystyle \frac{\left (3^{2x} \right )^{2}}{3^{8}}\\ 0&=\displaystyle \frac{\left (3^{2x} \right )^{2}}{3^{8}}-3\left ( 3^{2x} \right )\\ 0&=\left ( 3^{2x} \right )^{2}-3^{9}.\left ( 3^{2x} \right )\\ 0&=\left (3^{2x} \right )\left ( \left ( 3^{2x} \right )-3^{9} \right ) \end{aligned} &\color{black}\begin{aligned}\textbf{atau}\qquad\qquad\qquad&\\ \left ( 3^{2x} \right )^{2}-27.\left ( 3^{2x} \right )&=0\\ \left (3^{2x} \right )\left ( \left ( 3^{2x} \right )-3^{9} \right )&=0\\ 3^{2x}=0\: \: \textrm{atau}\: \: 3^{2x}&=3^{9}\\ (\textrm{tm})\quad \textrm{atau}\: \: 3^{2x}&=3^{9}\\ 2^{x}&=9\\ x&=\displaystyle \frac{9}{2}\\ \textrm{atau}\: \: x&=4\displaystyle \frac{1}{2} \end{aligned} \\\hline \end{array} \end{array}.

\color{red}\begin{array}{ll}\\ \fbox{16}.&\textrm{Jika diketahui}\: \: x^{x^{x^{x^{\cdots }}}}=2019,\\ &\textrm{maka nilai dari}\: \: x=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&\sqrt{2019}&&&\textrm{D}.&\sqrt{2019}^{\sqrt{2019}}\\ \textrm{B}.&\sqrt[2019]{2019}&\textrm{C}.&2019^{\sqrt{2019}}&\textrm{E}.&\sqrt{2019\sqrt{2019}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui}\\x^{x^{x^{x^{\cdots }}}}&=2019\\ \textrm{maka},&\\ x^{2019}&=2019\\ x&=\sqrt[2019]{2019} \end{aligned} \end{array}.

\color{red}\begin{array}{ll}\\ \fbox{17}.&\textrm{Jika}\: \: \sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{a}+\sqrt{b}+\sqrt{c},\\ &\textrm{maka nilai dari}\: \: \left ( a+b-c \right )^{abc}=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&1000&&&\textrm{D}.&-1\\ \textrm{B}.&1&\textrm{C}.&0&\textrm{E}.&-1000 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\begin{aligned}\left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )^{2}&=\left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )\times \left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )\\ &=\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.5}+\sqrt{3.2}+\sqrt{3.3}\\ &\qquad\qquad+\sqrt{3.5}+\sqrt{5.2}+\sqrt{5.3}+\sqrt{5.5}\\ &=2+2\sqrt{6}+2\sqrt{10}+3+2\sqrt{15}+5\\ &=2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\\ &=10+\sqrt{2^{2}.6}+\sqrt{2^{2}.10}+\sqrt{2^{2}.15}\\ &=10+\sqrt{24}+\sqrt{40}+\sqrt{60}\\ \sqrt{2}+\sqrt{3}+\sqrt{5}&=\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\\ &\begin{cases} a &=2 \\ b & =3 \\ c &=5 \end{cases}.\qquad \textrm{sesuai dengan urutannya}\\ \textrm{Sehingga nilai}\qquad &\\ \left ( a+b-c \right )^{abc}&=\left ( 2+3-5 \right )^{2.3.5}\\ &=0^{30}\\ &=0 \end{aligned} \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{18}.&\textrm{Nilai}\: \: \sqrt[3]{\displaystyle \frac{16}{\displaystyle \sqrt[3]{\frac{16}{\displaystyle \sqrt[3]{\displaystyle \frac{16}{...}}}}}}=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&8&&&\textrm{D}.&\displaystyle \frac{1}{4}\\ \textrm{B}.&4&\textrm{C}.&2&\textrm{E}.&\displaystyle \frac{1}{16} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\displaystyle \sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{...}}}}}}&=....\\ \textrm{misalkan}&\\ x&=\displaystyle \sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{...}}}}}}\\ x&=\sqrt[3]{\displaystyle \frac{16}{x}},\qquad \textrm{masing-masing ruas dipangkatkan 3, sehingga}\\ x^{3}&=\displaystyle \frac{16}{x}\\ x^{3}.x&=16\\ x^{4}&=16\\ x&=\sqrt[4]{16}=\sqrt[4]{2^{4}}\\ &=2\\ \therefore \displaystyle \sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{...}}}}}}&=2 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{19}.&\textrm{Jumlah semua nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\: \: 2019^{x^{2}-7x+7}=2020^{x^{2}-7x+7}\\ &\begin{array}{lllllllll}\\ \textrm{A}.&-7&&&\textrm{D}.&5\\ \textrm{B}.&-5&\textrm{C}.&-3&\textrm{E}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\begin{aligned}\begin{aligned}2019^{x^{2}-7x+7}&=2020^{x^{2}-7x+7}\\ \textrm{karena bas}&\textrm{is tidak sama}\\ \textrm{maka pang}&\textrm{kat haruslah}=0,\: \textrm{yaitu}:\\ x^{2}-7x+7&=0,\: \: \textrm{jumlah semua nilai yang memenuhi adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}\\ \textrm{karena}\qquad&x^{2}-7x+7=0\begin{cases} \textrm{akar-akarnya yaitu}\begin{cases} x_{1} \\ x_{2} \end{cases} \\ \textrm{koefisien} /\textrm{konstan}\begin{cases} a=1 \\ b=-7 \\ c=7 \end{cases} \end{cases}\\ \therefore \: \: x_{1}+x_{2}&=-\displaystyle \frac{-7}{1}\\ &=7 \end{aligned} \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{20}.&\textrm{Nilai eksak dari}\\ &\displaystyle \frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+\frac{1}{10^{-2018}+1}+...+\displaystyle \frac{1}{10^{2018}+1}+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1} \\ &\begin{array}{lllllllll}\\ \textrm{A}.&2020&&&\textrm{D}.&2021,5\\ \textrm{B}.&2020,5&\textrm{C}.&2021&\textrm{E}.&2022 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}\textrm{Misal},\: \: x&=\frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+\frac{1}{10^{-2018}+1}+...+\frac{1}{10^{2018}+1}+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ \textrm{maka},\, \: x&=\frac{1}{\frac{1}{10^{2020}}+1}+\frac{1}{\frac{1}{10^{2019}}+1}+\frac{1}{\frac{1}{10^{2018}}+1}+...+\frac{1}{10^{2018}+1}+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{10^{2019}}{1+10^{2019}}+\frac{10^{2018}}{1+10^{2018}}+...+\frac{1}{10^{2018}+1}+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{1}{10^{2020}+1}+\frac{10^{2019}}{1+10^{2019}}+\frac{1}{10^{2019}+1}+\frac{10^{2018}}{1+10^{2018}}++\frac{1}{10^{2018}+1}...+\frac{1}{1^{0}+1}\\ &=\frac{10^{2020}+1}{10^{2020}+1}+\frac{10^{2019}+1}{10^{2019}+1}+\frac{10^{2018}+1}{10^{2018}+1}+...+\frac{10^{1}+1}{10^{1}+1}+\frac{1}{1^{0}+1}\\ &=\underset{\textrm{sebanyak}\: 2020}{\underbrace{1+1+1+1+1+...+1}}+\frac{1}{2}\\ &=2020,5 \end{aligned} \end{array}.

Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Peminatan

\color{blue}\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai}\: \: m-n^{m-n}\: \: \textrm{untuk}\: \:m=2\: \: \textrm{dan}\: \: n=-2\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{A}.&-18&&&\textrm{D}.&18\\ \textrm{B}.&-14&\textrm{C}.&14&\textrm{E}.&256 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}m-n^{m-n}&=(2-(-2))^{2-(-2)}\\ &=4^{4}\\ &=256 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{2}.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{3^{m+8}\left (7^{3m+1}\times 3^{4m-1} \right )^{2}}{(7^{2m}\times 3^{3m+2})^{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{A}.&49&&&\textrm{D}.&7^{2m+2}\\ \textrm{B}.&9&\textrm{C}.&7&\textrm{E}.&3^{2m-1} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\displaystyle \frac{3^{m+8}\left (7^{3m+1}\times 3^{4m-1} \right )^{2}}{(7^{2m}\times 3^{3m+2})^{3}}&=\displaystyle \frac{3^{m+8+2(4m-1)}\times 7^{2(3m+1)}}{7^{3.2m}\times 3^{3(3m+2)}}\\ &=\displaystyle \frac{3^{m+8m+8-2}\times 7^{6m+2}}{3^{9m+6}\times 7^{6m}}\\ &=3^{0}\times 7^{2}\\ &=1\times 49\\ &=49 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{3}.&\textrm{Jika}\: \: \displaystyle \frac{(0,24)^{3}(0,243)^{5}}{(3,6)^{7}}=\frac{2^{b}3^{m}}{5^{k}}\:, \: \textrm{nilai}\: \: b+m+k\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{A}.&7&&&\textrm{D}.&10\\ \textrm{B}.&8&\textrm{C}.&9&\textrm{E}.&11 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\displaystyle \frac{(0,24)^{3}(0,243)^{5}}{(3,6)^{7}}&=\displaystyle \frac{\left (\frac{24}{100} \right )^{3}\times \left ( \frac{243}{1000} \right )^{5}}{\left (\frac{36}{10} \right )^{7}}\\ &=\displaystyle \frac{(8\times 3)^{3}\times \left (3^{5} \right )^{5}}{(2^{2}\times 3^{2})^{7}}\times \displaystyle \frac{10^{7}}{100^{3}\times 1000^{5}}\\ &=\displaystyle \frac{(2^{3}\times 3)^{3}\times 3^{25}}{(2^{14}\times 3^{14})}\times \displaystyle \frac{10^{7}}{\left ( 10^{2} \right )^{3}\times \left ( 10^{3} \right )^{5}}\\ &=2^{9-14}.3^{3+25-14}.10^{7-6-15}\\ &=2^{-5}.3^{14}.10^{-14}=2^{-5}.3^{14}.(2.5)^{-14}\\ &=2^{-5-14}.3^{21}.5^{-14}\\ &=\displaystyle \frac{2^{-19}.3^{14}}{5^{14}}\\ \textrm{Sehingga}\: \: &b+m+k=-19+14+14=9 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Perhatikanlah grafik berikut} \end{array}

\color{blue}\begin{array}{ll}\\ .\quad\: \, &\textrm{Rumus fungsi untuk grafik tersebut di ata adalah}\, ....\\ &\begin{array}{llllllll}\\ \textrm{A}.&f(x)=1+3^{2x-1}&&&\textrm{D}.&f(x)=-1+3^{2x-1}\\ \textrm{B}.&f(x)=1-3^{2x-1}&\textrm{C}.&f(x)=1-\left ( \displaystyle \frac{1}{3} \right )^{2x-1}&\textrm{E}.&f(x)=-1+\left ( \displaystyle \frac{1}{3} \right )^{2x-1} \end{array}\\\\ &\textrm{Jawab}:\\ &\textrm{kita uji titik-titiknya, misal di titik}\: \: (1,2)\: \: \textrm{dan}\: \: \left (\displaystyle \frac{1}{2},0 \right )\\ &\begin{aligned}\textrm{A}&\Rightarrow (1,2)\Rightarrow f(1)=1+3^{2.1-1}=1+3=4\neq 2\: \: (\textrm{salah})\\ \textrm{B}&\Rightarrow (1,2)\Rightarrow f(1)=1-3^{2.1-1}=1-3=-2\neq 2\: \: (\textrm{salah})\\ \textrm{C}&\Rightarrow (1,2)\Rightarrow f(1)=1-\left ( \displaystyle \frac{1}{3} \right )^{2.1-1}=1-\frac{1}{3}=\frac{2}{3}\neq 2\: \: (\textrm{salah})\\ \textrm{D}&\Rightarrow (1,2)\Rightarrow f(1)=-1+3^{2.1-1}=-1+3=2= 2\: \: (\textbf{benar})\\ \textrm{E}&\Rightarrow (1,2)\Rightarrow f(1)=-1+\left ( \displaystyle \frac{1}{3} \right )^{2.1-1}=-1+\frac{1}{3}=-\frac{2}{3}\neq 2\: \: (\textrm{salah})\\ \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{5}.&\textrm{Nilai}\: \: \displaystyle \frac{10^{2020}-10^{2019}}{9}\: \: \textrm{sama dengan}\: ....\\ &\begin{array}{llllll}\\ \textrm{A}.&\displaystyle \frac{1}{9}&&&\textrm{D}.&\displaystyle \frac{10^{2019}}{9}\\\\ \textrm{B}.&\displaystyle \frac{10}{9}&\textrm{C}.&10^{2014}&\textrm{E}.&10^{2019} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{E}\\ &\begin{aligned}\displaystyle \frac{10^{2020}-10^{2019}}{9}&=\displaystyle \frac{10^{2019}\left ( 10^{1}-1 \right )}{9}\\ &=10^{2019}.\displaystyle \frac{9}{9}\\ &=10^{2019} \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{6}.&\textrm{Jika}\: \: \displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}=4\: ,\: \textrm{maka}\: \: x=\: ....\\ &\begin{array}{llllll}\\ \textrm{A}.&\displaystyle 1,1&&&\textrm{D}.&\displaystyle 1,4\\\\ \textrm{B}.&\displaystyle 1,2&\textrm{C}.&1,3&\textrm{E}.&1,5\\\\ &&&&&(\textbf{\textit{SAT Test Math Level 2}})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}&=4\\ \left (3.x^{^{^{^{ \frac{3}{2}}}}} \right )^{2}&=4^{2}\\ 3^{2}.x^{3}&=4^{2}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\times \frac{3}{3}\\ x^{3}&\leq \displaystyle \frac{4^{2}}{3^{2}}\times \frac{4}{3}\\ x^{3}&\leq \left ( \displaystyle \frac{4^{3}}{3^{3}} \right )\\ x^{3}&\leq \left ( \displaystyle \frac{4}{3} \right )^{3}\\ x&\leq \displaystyle \frac{4}{3}\\ x&\leq 1,\overline{333}\\ x&\approx 1,3 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{7}.&\textrm{Nilai}\: \: \left ( -\displaystyle \frac{1}{16} \right )^{^{^{\frac{2}{3}}}}=\: ....\\ &\begin{array}{llllll}\\ \textrm{A}.&\displaystyle -0,25&&&\textrm{D}.&\displaystyle 0,35\\\\ \textrm{B}.&\displaystyle -0,16&\textrm{C}.&0,16&\textrm{E}.&\textrm{nilainya tidak real}\\\\ &&&&&(\textbf{\textit{SAT Test Math Level 2}})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\left ( -\displaystyle \frac{1}{16} \right )^{^{^{\frac{2}{3}}}}&=\sqrt[3]{\left ( -\displaystyle \frac{1}{16} \right )^{2}}\\ &=\sqrt[3]{\displaystyle \frac{1}{256}}\\ &=\sqrt[3]{\displaystyle \frac{1}{64\times 4}}\\ &=\sqrt[3]{\displaystyle \frac{1}{64}}\times \sqrt[3]{\displaystyle \frac{1}{4}}\\ &=\displaystyle \frac{1}{4}\times \sqrt[3]{\displaystyle \frac{1}{4}\times \frac{2}{2}}\\ &=\displaystyle \frac{1}{4}\times \sqrt[3]{\displaystyle \frac{1}{8}}\times \sqrt[3]{2}\\ &=\displaystyle \frac{1}{4}\times \frac{1}{2}\times \sqrt[3]{2}\\ &=\displaystyle \frac{1}{8}\times \sqrt[3]{2}\\ &=(0,125)\times (1,...)\\ &\approx 0,16 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{8}.&\textrm{Jika bentuk}\: \: \displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\: \: \textrm{dinyatakan dalam pangkat positif}=\: ....\\ &\begin{array}{llllll}\\ \textrm{A}.&\displaystyle \frac{a^{2}}{a-b}&&&\textrm{D}.&\displaystyle \frac{a^{2}}{b-a}\\\\ \textrm{B}.&\displaystyle \frac{a^{2}}{a-1}&\textrm{C}.&\displaystyle \frac{b-a}{ab}&\textrm{E}.&\displaystyle \frac{1}{a-b}\\\\ &&&&&(\textbf{\color{black}\textit{SAT Test Math Level 2}})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\begin{aligned}\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}&=\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}\\ &=\displaystyle \frac{a^{2}}{b-a} \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{9}.&\textrm{Jika}\: \: \displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12}\: ,\\\\ &\textrm{maka}\: \: a+b+c=....\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textbf{\color{black}\textit{UM UGM 2016 Mat Das}})\\ &\begin{array}{llllllll}\\ \textrm{A}.&0&&&\textrm{D}.&3\\ \textrm{B}.&1&\textrm{C}.&2&\textrm{E}.&4 \end{array}\\\\ &\textrm{Jawab}:\qquad \textbf{E}\\ &\begin{aligned}\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}&=\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times \displaystyle \frac{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left ( \sqrt{2}+\sqrt{3} \right )^{2}-\left ( \sqrt{5} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times \displaystyle \frac{\sqrt{6}}{\sqrt{6}}\\ &=\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}\\ &=\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}\\ &=\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}\\ &\quad \begin{cases} a &=3 \\ b &=2 \\ c &=-1 \end{cases}\\ a+b+c&=3+2+(-1)\\ &=4 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{10}.&\textrm{Bentuk sederhana dari}\: \: \sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textbf{\color{black}\textit{SIMAK UI 2012 Mat IPA}})\\ &\begin{array}{llllllll}\\ \textrm{A}.&2-\sqrt{2}&&&\textrm{D}.&2+5\sqrt{2}\\ \textrm{B}.&8-\sqrt{2}&\textrm{C}.&-2+\sqrt{2}&\textrm{E}.&8+5\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\textrm{misalkan},\\ &\color{red}\begin{aligned}x&=\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}\\ &=\left ( \sqrt{33+20\sqrt{2}}-\sqrt{27-2.9\sqrt{2}}\: \right )\\ &=\sqrt{33+20\sqrt{2}}-\sqrt{27-18\sqrt{2}}\\ x^{2}&=33+20\sqrt{2}+27-18\sqrt{2}-2\sqrt{\left ( 33+20\sqrt{2} \right )\left ( 27-18\sqrt{2} \right )}\\ &=60+2\sqrt{2}-2\sqrt{33.27-33.18\sqrt{2}+27.20\sqrt{2}-20.18.2}\\ &=60+2\sqrt{2}-2\sqrt{891-720+540\sqrt{2}-594\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-54\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2.27\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{27.27}\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\sqrt{162+9-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\left ( \sqrt{162}-\sqrt{9} \right )\\ &=60+2\sqrt{2}-2\left ( 9\sqrt{2}-3 \right )\\ x^{2}&=66-16\sqrt{2}\\ x&=\sqrt{66-2.8\sqrt{2}}\\ &=\sqrt{64+2-2\sqrt{64.2}}\\ &=\sqrt{64}-\sqrt{2}\\ &=8-\sqrt{2} \end{aligned} \end{array}.

Lanjutan Contoh Soal Trigonometri

\color{blue}\begin{array}{ll}\\ 3.&\textrm{Sederhanakanlah bentuk berikut ini}\\ &\begin{array}{ll}\\ \textrm{a}.&\sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )\\ \textrm{b}.&\sin \left ( x+\displaystyle \frac{1}{4}m \right )+\sin \left ( x-\displaystyle \frac{1}{4}m \right )\\ \textrm{c}.&\cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )\\ \textrm{d}.&\cos \left ( \displaystyle \frac{5}{4}m+3x \right )+\cos \left ( \displaystyle \frac{5}{4}m-3x \right ) \end{array} \end{array}.

Solusi

\begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad \sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )&=2\cos \displaystyle \frac{\left (2x+60^{\circ} \right )+\left ( 2x-60^{\circ} \right )}{2}\sin \displaystyle \frac{\left (2x+60^{\circ} \right )-\left ( 2x-60^{\circ} \right )}{2}\\ &=2\cos \displaystyle \frac{4x}{2}\sin \displaystyle \frac{120^{\circ}}{2}\\ &=2\cos 2x\sin 60^{\circ}\\ &=2\cos 2x.\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\sqrt{3}\cos 2x \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad \cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )&=-2\sin \displaystyle \frac{\left (2x+4y \right )+\left ( 2x-4y \right )}{2}\sin \displaystyle \frac{\left (2x+4y \right )-\left ( 2x-4y \right )}{2}\\ &=-2\sin \displaystyle \frac{4x}{2}\sin \displaystyle \frac{8y}{2}\\ &=-2\sin 2x\sin 4y\\ \end{aligned} \\\hline \end{array}.