Contoh Soal Persiapan Semester Gasal Peminatan Kelas X (K13 Revisi)

\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai}\: \: \displaystyle \frac{10^{2018}-10^{2017}}{9}\: \: \textrm{sama dengan}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{1}{9}&&&\textrm{d}.&\displaystyle \frac{10^{2017}}{9}\\\\ \textrm{b}.&\displaystyle \frac{10}{9}&\textrm{c}.&10^{2012}&\textrm{e}.&10^{2017} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\displaystyle \frac{10^{2018}-10^{2017}}{9}&=\displaystyle \frac{10^{2017}\left ( 10^{1}-1 \right )}{9}\\ &=10^{2017}.\displaystyle \frac{9}{9}\\ &=10^{2017} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\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}.

\begin{array}{ll}\\ \fbox{3}.&\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}.

\begin{array}{ll}\\ \fbox{4}.&\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{\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}.

\begin{array}{ll}\\ \fbox{5}.&\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{\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}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Bentuk sederhana dari}\: \: \sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textbf{\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},\\ &\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}.

\begin{array}{ll}\\ \fbox{7}.&\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\qquad\qquad\qquad (\textbf{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}\\ &=0 \end{aligned} \end{array}.

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Memperingati Maulidur Rosul Muhammad S.A.W 2017

Allahumma sholli ‘alaa sayyidinaa Muhammad

Dengan memperingati maulid Nabi Muhammad S.A.W semoga kita dapat meneladi ahlaqul karimah beliau, karena beliaulah sebaik-baik uswatun hasanah.

Berikut biografi baginda Rosulullah Muhammad S.A.W yang disampaikan oleh salah seorang ustad kondang Ust. Abdul Somad Lc., M.A.

 

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Lanjutan Contoh Soal Persiapan Semester Gasal Kelas X (K13 Revisi)

\begin{array}{ll}\\ \fbox{11}.&\textrm{Diketahui grafik fungsi}\: \: f(x)=2x^{2}-ax+2\: \: \textrm{definit positif jika interval}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&a>-4&&&\textrm{d}.&4<a<6\\ \textrm{b}.&a>4&\textrm{c}.&-4<a<4&\textrm{e}.&-6<a<4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diketahui fungsi kuadrat}&:\\ f(x)&=2x^{2}-ax+2\begin{cases} A &=2 \\ B &=-a \\ C &=2 \end{cases}\\ \textrm{Syarat \textbf{definit positif}}&\: \textrm{adalah}:\\ &\begin{cases} a &>0,\quad \textrm{jelas}\quad a=2>0 \\ D &<0 \qquad \Rightarrow D=B^{2}-4AC \end{cases}\\ \textrm{Tinggal kondisi}\: &\: D-\textrm{nya}\\ D&=B^{2}-4AC\\ (-a)^{2}-4(2)(2)&<0\\ a^{2}-16&<0\\ (a+4)(a-4)&<0\\ \textrm{Berikut garis}&\: \textrm{bilangannya}:\\ &\begin{array}{lll|llll|llll} &\multicolumn{3}{c}{.}&&&\multicolumn{3}{c}{.}&\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{3}{l}{.}&&\\\cline{4-7} +&+&&-&-&-&&&+&+\\\hline &&\multicolumn{2}{l}{\textcircled{-4}}&&&\multicolumn{2}{l}{\textcircled{4}}&&\multicolumn{2}{r}{\textbf{sumbu}-a}\\ \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Jika}\: \: 0<m<10\: ,\: \textrm{maka sifat dari fungsi kuadrat}\: \: f(x)=mx^{2}+2mx+10\\ & \textrm{berikut yang memenuhi adalah}....\\ &\begin{array}{ll}\\ \textrm{a}.&\textrm{selalu negatif}\\ \textrm{b}.&\textrm{selalu positif}\\ \textrm{c}.&\textrm{hanya positif di setiap}\: x\: \textrm{dengan}\: 0<x<10\\ \textrm{d}.&\textrm{hanya negatif di setiap}\: x\: \textrm{dengan}\: 0<x<10\\ \textrm{e}.&\textrm{hanya positif di setiap}\: x\: \textrm{dengan}\: x<0\: \: \textrm{atau}\: \: x>10\\\\ &\qquad\qquad\qquad (\textbf{SBMPTN 2014 Mat Das}) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}f(x)&=mx^{2}+2mx+10\\ &\qquad \begin{cases} a &=m \\ b &=2m \\ c &= 10 \end{cases}\\ D&=b^{2}-4ac=(2m)^{2}-4(m)(10)\\ &=4m^{2}-40m\\ &=4m(m-10)\\ \textrm{dik}&\textrm{etahui dari \textbf{soal} bahwa}\: \: 0<m<10\\ \textrm{kar}&\textrm{ena}\: \: \begin{cases} a &=m\: \: \textrm{dengan} =0<m<10 \Rightarrow \textbf{berarti \textit{a} semuanya positif}\\ D & \textrm{untuk}\: =4m(m-10)\Rightarrow \textbf{nilai \textit{D} akan selalu negatif} \end{cases}\\ \textrm{ma}&\textrm{ka fungsi}\: \: f(x)\: \: \textrm{pada selang}\: \: 0<m<10\: \: \textrm{akan \textbf{definit positif}}\\ \textrm{jan}&\textrm{gan terkecoh dengan jawaban c}\: ,\: \textbf{jawaban \textit{c} akan salah misal saja saat x=0} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Himpunan penyelesaian untuk}\: \: \displaystyle \frac{2x}{1-x}< 3\: \: \textrm{adalah}....\\ &\begin{array}{ll}\\ \textrm{a}.&\left \{ x\in \mathbb{R}\mid x<\displaystyle \frac{3}{5} \right \}\\\\ \textrm{b}.&\left \{ x\in \mathbb{R}\mid x<\displaystyle \frac{5}{3} \right \}\\\\ \textrm{c}.&\left \{ x\in \mathbb{R}\mid x>\displaystyle \frac{3}{5} \right \}\\\\ \textrm{d}.&\left \{ x\in \mathbb{R}\mid \displaystyle \frac{3}{5}<x<1 \right \}\\\\ \textrm{e}.&\left \{ x\in \mathbb{R}\mid x<\displaystyle \frac{3}{5}\: \: \textrm{atau}\: \: x>1\right \}\\\\ &\qquad\qquad\qquad (\textbf{SBMPTN 2014 Mat Das}) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\displaystyle \frac{2x}{1-x}&< 3\\ \displaystyle \frac{2x}{1-x}- 3&<0\\ \displaystyle \frac{2x-3(1-x)}{1-x}&< 0\\ \displaystyle \frac{5x-3}{1-x}&< 0\quad \textrm{dikalikan dengan}\: \: (-1)\\ \displaystyle \frac{5x-3}{x-1}&> 0 \quad \textrm{senilai dengan}\: \: (5x-3)(x-1)>0\\ \textrm{garis}&\: \textrm{bilangannya}\\ &\begin{array}{lll|llll|llll} &\multicolumn{3}{c}{.}&&&\multicolumn{3}{c}{.}&\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{3}{l}{.}&&\\\cline{1-3}\cline{8-10} +&+&&-&-&-&&&+&+\\\hline &&\multicolumn{2}{l}{\displaystyle \frac{3}{5}}&&&\multicolumn{2}{l}{1}&&\multicolumn{2}{r}{\textbf{sumbu}-x}\\ \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Himpunan penyelesaian untuk}\: \: \displaystyle \frac{x-2}{2x+3}< 1\: \: \textrm{senilai dengan}\: \: \left | 4x+a \right |>b.\\ &\textrm{dengan nilai}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{berturut-turut adalah}....\\ &\begin{array}{ll}\\ \textrm{a}.&7\: \: \textrm{dan}\: \: 13\\ \textrm{b}.&13\: \: \textrm{dan}\: \: 7\\ \textrm{c}.&6\: \: \textrm{dan}\: \: 13\\ \textrm{d}.&13\: \: \textrm{dan}\: \: -6\\ \textrm{e}.&-13\: \: \textrm{dan}\: \: 7\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\displaystyle \frac{x-2}{2x+3}&< 1\\ \displaystyle \frac{x-2}{2x+3}- 1&<0\\ \displaystyle \frac{x-2-1(2x+3)}{2x+3}&< 0\\ \displaystyle \frac{-x-5}{2x+3}&< 0\quad \textrm{dikalikan dengan}\: \: (-1)\\ \displaystyle \frac{x+5}{2x+3}&> 0 \quad \textrm{senilai dengan}\: \: (x+5)(2x+3)>0\\ \textrm{garis}&\: \textrm{bilangannya}\\ &\begin{array}{lll|llll|llll} &\multicolumn{3}{c}{.}&&&\multicolumn{3}{c}{.}&\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{3}{l}{.}&&\\\cline{1-3}\cline{8-10} +&+&&-&-&-&&&+&+\\\hline &&\multicolumn{2}{l}{-5}&&&\multicolumn{2}{l}{-\displaystyle \frac{3}{2}}&&\multicolumn{2}{r}{\textbf{sumbu}-x}\\ \end{array} \end{aligned} &\begin{aligned}\left | 4x+a \right |&>b\\ \left ( 4x+a \right )^{2}&>b^{2}\\ \left ( 4x+a \right )^{2}-b^{2}&>0\\ (4x+a+b)(4x+a-b)&>0\\ \textrm{coba perhatikan}&\\ a+b=20\: \: \textrm{dan}\: \: a-b&=12 \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Himpunan penyelesaian}\: \: \: \displaystyle \frac{x^{2}-3|x|+2}{x+1}\geq 0\: \: \textrm{ adalah}....\\ &\begin{array}{ll}\\ \textrm{a}.&\left \{ x\in \mathbb{R}\mid -1<x\leq 1\: \: \textrm{atau}\: \: x\geq 2 \right \}\\ \textrm{b}.&\left \{ x\in \mathbb{R}\mid -2<x< -1\: \: \textrm{atau}\: \: x\geq 2 \right \}\\ \textrm{c}.&\left \{ x\in \mathbb{R}\mid -2\leq x< -1\: \: \textrm{atau}\: \: -1<x\leq 1 \right \}\\ \textrm{d}.&\left \{ x\in \mathbb{R}\mid -2\leq x< -1\: \: \textrm{atau}\: \: -1<x\leq 1\: \: \textrm{atau}\: \: x\geq 2\right \}\\ \textrm{e}.&\left \{ x\in \mathbb{R}\mid x<-1\: \: \textrm{atau}\: \: -1<x\leq 1\: \: \textrm{atau}\: \: x\geq 2 \right \}\\\\ &\qquad\qquad\qquad\qquad\qquad\qquad (\textbf{SIMAK UI 2011 Mat IPA}) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{array}{|c|c|}\hline x\geq 0&x<0\\\hline \begin{aligned}&\\ \displaystyle \frac{x^{2}-3|x|+2}{x+1}&\geq 0\\ \displaystyle \frac{x^{2}-3x+2}{x+1}&\geq 0\\ \displaystyle \frac{(x-1)(x-2)}{x+1}&\geq 0\qquad (\textbf{mm})\\ \textrm{garis}&\: \textrm{bilangannya}\\ &\begin{array}{ll|l|l|llll|llll} \multicolumn{3}{l}{.}&\multicolumn{3}{c}{.}&&\multicolumn{3}{c}{.}&\\\cline{4-11} \multicolumn{3}{l|}{.}&\multicolumn{2}{l}{.}&&&\multicolumn{3}{l}{.}&&\\\cline{3-4}\cline{9-11} -& -&+\: \: \: +&+&-&-&-&&&+&+\\\hline \multicolumn{2}{r}{-1}&\multicolumn{1}{r}{\qquad\textcircled{0}}&\multicolumn{2}{c}{ \textcircled{1}}&&&\multicolumn{2}{c}{\textcircled{2}}&&\multicolumn{2}{r}{\textbf{sumbu}-x}\\ \end{array}\\ & \end{aligned} &\begin{aligned}\displaystyle \frac{x^{2}-3|x|+2}{x+1}&\geq 0\\ \displaystyle \frac{x^{2}-3(-x)+2}{x+1}&\geq 0\\ \displaystyle \frac{x^{2}+3x+2}{x+1}&\geq 0\\ \displaystyle \frac{(x+1)(x+2)}{x+1}&\geq 0\\ (x+2)&\geq 0\\ x&\geq 2 \qquad (\textbf{tm}) & \end{aligned}\\\hline \end{array} \end{array}.

Sebagai catatan : antara  x = -1 sampai x = 0, gunakanlah sembarang titik uji.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Himpunan penyelesaian untuk}\: \: \displaystyle \left | x-2 \right |-1\geq x\: \: \textrm{adalah}....\\ &\begin{array}{ll}\\ \textrm{a}.&\left \{ x\mid 0\leq x<\displaystyle \frac{7}{2} \right \}\\\\ \textrm{b}.&\left \{ x\mid x\geq 0 \right \}\\\\ \textrm{c}.&\left \{ x\mid x\leq \displaystyle \frac{1}{2} \right \}\\\\ \textrm{d}.&\left \{ x\mid 0\leq x< \displaystyle \frac{5}{2} \right \}\\\\ \textrm{e}.&\left \{ x\mid -1\leq x<\displaystyle \frac{1}{2}\right \}\\\\ &\qquad\qquad\qquad (\textbf{UM UGM 2013 Mat IPA}) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{array}{|c|c|}\hline x\geq 2&x< 2\\\hline \begin{aligned}\left | x-2 \right |-1&\geq x\\ (x-2)-1&\geq x\\ (x-2)-1-x&\geq 0\\ -3&\geq 0\quad (\textbf{tm})\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\left | x-2 \right |-1&\geq x\\ -(x-2)-1&\geq x\\ -x+2-1-x&\geq 0\\ -2x+1&\geq 0\\ 2x-1&\leq 0\\ 2x&\leq 1\\ x&\leq \displaystyle \frac{1}{2}\quad (\textbf{mm})\\ &\begin{array}{lll|lll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&&\\\cline{1-6} &\multicolumn{2}{c}{.}&&&&&&\\\cline{1-3} &&&&&&&&\\\hline &&\multicolumn{2}{c}{(\frac{1}{2})}&&\multicolumn{2}{c}{2}&\\ \end{array} \\ & \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Rentang nilai yang memenuhi}\: \: x+2> \sqrt{10-x^{2}}\: \: \textrm{adalah}....\\ &\begin{array}{ll}\\ \textrm{a}.&-\sqrt{10}\leq x\leq \sqrt{10}\\ \textrm{b}.&x<-3\: \: \textrm{atau}\: \: x>1\\ \textrm{c}.&2\leq x\leq \sqrt{10}\\ \textrm{d}.&1<x\leq \sqrt{10}\\ \textrm{e}.&-3< x\leq \sqrt{10}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textbf{Syarat Daerah Asal (SDA)}}\\\hline \begin{aligned}x+2&\geq 0\\ x&>-2\\ &\begin{array}{lll|lllllll}\\ &\multicolumn{2}{c}{.}&&&&&&\\\cline{4-8} &&&&&&&&\\\hline &&\multicolumn{2}{c}{\textcircled{-2}}&&&&\\ \end{array}\\ &\\ &\\ & \end{aligned}&\begin{aligned}10-x^{2}&\geq 0\\ x^{2}-10&\leq 0\\ x^{2}&\leq 10\\ x&\leq \sqrt{10}\\ &\begin{array}{lll|lllllll}\\ &\multicolumn{2}{c}{.}&&&&&&\\\cline{1-3} &&&&&&&&\\\hline &&\multicolumn{2}{l}{\left (\sqrt{10} \right )}&&&&\\ \end{array}\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\textbf{Proses Pengkuadratan}}\\\hline \multicolumn{2}{|c|}{\begin{aligned}x+2&> \sqrt{10-x^{2}}\\ (x+2)^{2}&>\left (\sqrt{10-x^{2}} \right )^{2}\\ x^{2}+4x+4&> 10-x^{2}\\ 2x^{2}+4x-6&>0\\ &\begin{array}{lll|llll|llll} &\multicolumn{3}{c}{.}&&&\multicolumn{3}{c}{.}&\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{3}{l}{.}&&\\\cline{1-3}\cline{8-10} +&+&&-&-&-&&&+&+\\\hline &&\multicolumn{2}{l}{-3}&&&\multicolumn{2}{l}{1}&&\multicolumn{2}{r}{\textbf{sumbu}-x}\\ \end{array}\\ & \end{aligned}}\\\hline \end{array} \end{array}.

Silahkan digabung garis bilangannya, mana yang daerah yang mendapatkan arsiran tripel, itulah daerah penyelesaiannya.

\begin{array}{ll}\\ \fbox{18}.&\textrm{Daerah asal (domain) dari}\: \: f(x)=\sqrt{\displaystyle \frac{3x^{2}+x-2}{2x^{2}-5x+2}}\: \: \textrm{adalah}....\\ &\begin{array}{ll}\\ \textrm{a}.&x\leq -1\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}<x\leq \frac{2}{3}\: \: \textrm{atau}\: \: x>2\\\\ \textrm{b}.&x\leq -2\: \: \textrm{atau}\: \: \displaystyle \frac{1}{3}<x<\frac{2}{3}\: \: \textrm{atau}\: \: x>2\\\\ \textrm{c}.&x\leq -2\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x\neq 1\\\\ \textrm{d}.&-2\leq x<\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \displaystyle \frac{2}{3}\leq x<1\\\\ \textrm{e}.&x<-1\: \: \textrm{atau}\: \: x>2\\\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Diketahui bahwa}&\: \: f(x)=\sqrt{\displaystyle \frac{3x^{2}+x-2}{2x^{2}-5x+2}}\\ \textrm{maka domainnya}&\: \textrm{adalah}:\\ &\displaystyle \frac{3x^{2}+x-2}{2x^{2}-5x+2}\geq 0\\ &\displaystyle \frac{(x+1)(3x-2)}{(x-2)(2x-1)}\geq 0\\ &\begin{array}{lll|ll|lll|lll|lllllll}\\ &\multicolumn{2}{c}{.}&&\multicolumn{2}{c}{.}&\multicolumn{2}{c}{.}&&\multicolumn{2}{c}{.}&\\\cline{1-3}\cline{6-8}\cline{12-14} +&+&&&&+&+&+&&&&+&+&\\\hline &&\multicolumn{2}{c}{\textcircled{-1}}&\multicolumn{2}{c}{\displaystyle \frac{1}{2}}&&\multicolumn{2}{r}{\left (\displaystyle \frac{2}{3} \right )}&&\multicolumn{2}{c}{2}\\ \end{array} \end{aligned} \end{array}.

Perhatikan tanda pertidaksamaannya, bagian penyebut tidak boleh sama dengan nol.

\begin{array}{ll}\\ \fbox{19}.&\textrm{Tentukanlah interval nilai-nilai}\: \: m\: \: \textrm{sehingga}\: \: 9mx^{2}+24x+10\geq m\: \: \textrm{untuk setiap}\: \: x\: \: \textrm{real}\\\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textit{NUS Singapore Entrance Examination})\\\\ &\textrm{Jawab}:\\ &\begin{aligned}9mx^{2}+24x+10&\geq m\: \\ 9mx^{2}+24x+10-m&\geq 0\\ \textrm{ambil bagian yang}&\\ 9mx^{2}+24x+10-m&= 0\\ \textrm{Syarat setiap}\: \: x&\: \: \textrm{real adalah harga D}=A^{2}-4AC\geq 0\\ B^{2}-4AC&\geq 0\\ (24)^{2}-4\left ( 9m \right )\left ( 10-m \right )&\geq 0\\ 24^{2}-360m+36m^{2}&\geq 0\\ 16-10m+m^{2}&\geq 0\\ m^{2}-10m+16&\geq 0\\ (m-2)(m-8)&\geq 0\\ &\begin{array}{lll|llll|llll} &\multicolumn{3}{c}{.}&&&\multicolumn{3}{c}{.}&\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{3}{l}{.}&&\\\cline{1-3}\cline{8-10} +&+&&-&-&-&&&+&+\\\hline &&\multicolumn{2}{l}{\textcircled{2}}&&&\multicolumn{2}{l}{\textcircled{8}}&&\multicolumn{2}{r}{\textbf{sumbu}-x}\\ \end{array}\\\\ \textbf{HP}&=\left \{ m\mid m\leq 2\: \: \textrm{atau}\: \: m\geq 8,\: \:  m\in \mathbb{R} \right \} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{20}.&\textrm{Diberikan gambar daerah yang diarsir berikut ini}....\\ \end{array}\\.

.\qquad \, \: \begin{array}{l}\\ \textrm{Daerah yang diarsir merupakan daerah penyelesaian dari}\\ \textrm{a}.\quad x^{2}+y^{2}\leq 4\: ;\: x+2y\leq 4\\ \textrm{b}.\quad x^{2}+y^{2}\leq 4\: ;\: x+2y\geq 4\\ \textrm{c}.\quad x^{2}+y^{2}\leq 16\: ;\: y\geq x+1\\ \textrm{\textcircled{d}}.\quad x^{2}+y^{2}\leq 16\: ;\: y\leq x+1\\ \textrm{e}.\quad x^{2}+y^{2}\leq 16\: ;\: x-y\leq 1\\\\\\ \textbf{Jawabannya adalah opsi yang dilingkari yaitu d} \end{array}.

 

Sumber Referensi

  1. Kanginan, M,. 2017. Matematika untuk Siswa SMA-MA/SMA-MAK Kelas X (Cetakan ke-2). 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.
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Contoh Soal Persiapan Semester Gasal Kelas X (K13 Revisi)

\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi untuk}\: \: \left | 2x+5 \right |=9\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&2&&&\textrm{d}.&-7\: \: \textrm{dan}\: \: 2\\ \textrm{b}.&2\: \: \textrm{dan}\: \: 7&\textrm{c}.&-7\: \: \textrm{dan}\: \: -2&\textrm{e}.&-2\: \: \textrm{dan}\: \: 7 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\left | 2x+5 \right |&=9\\ \left ( 2x+5 \right )&=\pm 9\\ 2x&=\pm 9-5\\ x&=\displaystyle \frac{\pm 9-5}{2}\\ x&=\begin{cases} =\displaystyle \frac{+9-5}{2}=\frac{4}{2}=2 \\\\ =\displaystyle \frac{-9-5}{2}=\frac{-14}{2}=-7 \end{cases} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Penyelesaian yang memenuhi untuk}\: \: \left | 3x-(4x-7) \right |=6\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -13,-1 \right \}&&&\textrm{d}.&\left \{ -13,1 \right \}\\ \textrm{b}.&\left \{ -1,13 \right \}&\textrm{c}.&\left \{ 1,13 \right \}&\textrm{e}.&\left \{ 13 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\left | 3x-(4x-7) \right |&=6\\ \left | 3x-4x+7 \right |&=6\\ \left | -x+7 \right |&=6\\ \left ( -x+7 \right )&=\pm 6\\ -x+7&=\begin{cases} 6 \Leftrightarrow -x=6-7\Leftrightarrow x=1\\\\ -6\Leftrightarrow -x=-6-7\Leftrightarrow x=13 \end{cases} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Penyelesaian yang memenuhi untuk}\: \: \left | x-1 \right |=2x+1\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ -2,0 \right \}&\textrm{c}.&\left \{ -1 \right \}&\textrm{e}.&\left \{ 0 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\left | x-1 \right |&=2x+1\\ (x-1)&=\pm (2x+1)\\ (x-1)&= \begin{cases} +(2x+1) \\\\ -(2x+1) \end{cases}\\\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Syarat}&:\\ (x-1)&\begin{cases} x-1\geq 0 \Leftrightarrow x\geq 1\\\\ x-1<0  \Leftrightarrow x<1 \end{cases}\\ & \end{aligned}}\\\hline x\geq 1&x< 1\\\hline \multicolumn{2}{|c|}{\textrm{Proses}}\\\hline \begin{aligned}(x-1)&=+(2x+1)\\ x-2x&=1+1\\ -x&=2\\ x&=-2 \end{aligned}&\begin{aligned}(x-1)&=-(2x+1)\\ x+2x&=-1+1\\ 3x&=0\\ x&=0 \end{aligned}\\\hline \textrm{tidak memenuhi}&\textbf{memenuhi}\\\hline \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Penyelesaian yang memenuhi untuk}\: \: \left | 3x+1 \right |=2x+9\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ 8 \right \}&\textrm{c}.&\left \{ -2,8 \right \}&\textrm{e}.&\textrm{setiap bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left | 3x+1 \right |&=2x+9\\ (3x+1)&=\pm (2x+9)\\ (3x+1)&= \begin{cases} +(2x+9) \\\\ -(2x+9) \end{cases}\\\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Syarat}&:\\ (3x+1)&\begin{cases} 3x+1\geq 0 \Leftrightarrow x\geq -\displaystyle \frac{1}{3}\\\\ 3x+1<0 \Leftrightarrow x<-\displaystyle \frac{1}{3} \end{cases}\\ & \end{aligned}}\\\hline x\geq -\displaystyle \frac{1}{3}&x< -\displaystyle \frac{1}{3}\\\hline \multicolumn{2}{|c|}{\textrm{Proses}}\\\hline \begin{aligned}(3x+1)&=+(2x+9)\\ 3x-2x&=9-1\\ x&=8\\ & \end{aligned}&\begin{aligned}(3x+1)&=-(2x+9)\\ 3x+2x&=-9-1\\ 5x&=-10\\ x&=-2 \end{aligned}\\\hline \textbf{memenuhi}&\textbf{memenuhi}\\\hline \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Jumlah akar-akar dari}\: \: x^{2}+\left | x \right |-6=0\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-1&&&\textrm{d}.&2\\ \textrm{b}.&0&\textrm{c}.&1&\textrm{e}.&4\\\\ &&&&&(\textbf{Entrance Examination}) \end{array}\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}x^{2}+\left | x \right |-6&=0\\ (\left | x \right |+3)(\left | x \right |-2)&=0\\ \left | x \right |+3=0\quad \textrm{atau}\quad \left | x \right |-2&=0\\ \left | x \right |=-3\: (\textbf{tm})\quad \textrm{atau}\quad \left | x \right |&=2\: (\textbf{mm})\\ x&=\pm 2\begin{cases} x_{1}&=2 \\ x_{2} &=-2 \end{cases}\\\\ \textrm{untuk jumlah}&\: \textrm{dari akar-akarnya adalah}:\\ x_{1}+x_{2}&=2+(-2)\\ &=0 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\: \: \left | 2x-9 \right |< 3\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-3\leq x\leq 6&&&\textrm{d}.&3\leq x\leq 6\\ \textrm{b}.&-3<x<6&\textrm{c}.&3<x<6&\textrm{e}.&-3<x<-6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left | 2x-9 \right |&< 3\\ -3<2x-9&<3\\ -3+(9)<2x-9+(9)&<3+(9)&\textnormal{masing-masing ditambah 9}\\ 6<2x&<12\\ 6.\left ( \displaystyle \frac{1}{2} \right )<2x.\left ( \displaystyle \frac{1}{2} \right )&<12.\left ( \displaystyle \frac{1}{2} \right )&\textnormal{masing-masing dikali}\: \: \displaystyle \frac{1}{2}\\ 3<x&<6 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{7}.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\: \: \left | 3x+5 \right |\geq 19\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&x\leq -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x\geq 8&&&\textrm{d}.&x< -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x> 8\\ \textrm{b}.&x<-8\: \: \textrm{atau}\: \: x>\displaystyle \frac{14}{3}&\textrm{c}.&x\leq -8\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{14}{3}&\textrm{e}.&x\leq 8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{14}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\left | 3x+5 \right |\geq 19\\\\ (\ast )-19\geq 3x+5&\quad \textrm{atau}&(\ast \ast )\: \: 3x+5\geq 19\\ -19-5\geq 3x&\quad \textrm{atau}&3x\geq 19-5\\ -\displaystyle \frac{24}{3}\geq x&\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ -8\geq x&\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ x\leq -8&\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3} \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{8}.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi}\: \: 25-\left | 10x+5 \right |\geq \left | 40x-20 \right |\: \: \textrm{adalah}... .\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textit{NUS Entrance Examination A level})\\\\ &\textrm{Jawab}:\\ &\begin{aligned}25&-\left | 10x+5 \right |\geq \left | 40x-20 \right |\\ 25&-5\left | 2x+1 \right |\geq 20\left | 2x-1 \right |\\ 5&-\left | 2x+1 \right |\geq 4\left | 2x-1 \right |\\ \textrm{ilustrasinya}&\quad \begin{array}{llllllllll} &&&&&&\\\hline &&-\frac{1}{2}&&&\frac{1}{2}&& \end{array}\\ \textrm{dan berikut}&\: \textrm{pembagian wilayahnya}\\ &\begin{array}{|c|c|c|}\hline -\infty < x< -\displaystyle \frac{1}{2}&-\displaystyle \frac{1}{2}\leq x< \frac{1}{2}&\displaystyle \frac{1}{2}\leq x< \infty \\\hline \begin{cases} \left | 2x+1 \right | &=-(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=+(2x-1) \end{cases}\\\hline \end{array} \end{aligned} \end{array}..

Selanjutnya,

\begin{array}{|c|c|c|}\hline \begin{aligned}\textrm{\underline{Untuk}}:\: \: -\infty < x<& -\frac{1}{2}\: ,\\ 25-\left | 10x+5 \right |&\geq \left | 40x-20 \right |\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(-(2x+1))&\geq 4(-(2x-1))\\ 5+2x+1&\geq -8x+4\\ 10x&\geq -2\\ x&\geq -\frac{2}{10}\quad (\textbf{tm})\\ & \end{aligned}&\begin{aligned}\textrm{\underline{Untuk}}:\: \: -\frac{1}{2} \leq x<& \frac{1}{2}\: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1)&\geq 4(-(2x-1))\\ 5-2x-1&\geq -8x+4\\ 6x&\geq 0\\ x&\geq 0\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ 0\leq x< \frac{1}{2} \right \} \end{aligned} &\begin{aligned}\textrm{\underline{Untuk}}:\: \: \frac{1}{2}\leq x< &\infty \: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1))&\geq 4(2x-1)\\ 5-2x-1&\geq 8x-4\\ -10x&\geq -8\\ x&\leq \frac{8}{10}\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ \frac{1}{2}\leq x\leq \frac{4}{5} \right \} \end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Sehingga yang memenuhi}&\: \textrm{adalah}:\\ &=\left \{ 0\leq x\leq \displaystyle \frac{4}{5} \right \} &\\ & \end{aligned}}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: (x+3)(x-1)\geq (x-1)\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&1\leq x\leq 3&&&\textrm{d}.&-2\geq x\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{b}.&x\leq -2\: \: \textrm{atau}\: \: x\geq 1&\textrm{c}.&-3\leq x\leq -1&\textrm{e}.&-1\geq x\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}(x+3)(x-1)&\geq (x-1)\\ (x+3)(x-1)-(x-1)&\geq 0\\ (x-1)\left ( (x+3)-1 \right )&\geq 0\\ (x-1)(x+2)&\geq 0\\ \textrm{Berikut garis bilangan}&\textrm{nya}\\ &\begin{array}{lll|llll|llll} &\multicolumn{3}{c}{.}&&&\multicolumn{3}{c}{.}&\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{3}{l}{.}&&\\\cline{1-3}\cline{8-9} +&+&&-&-&-&&&+&+\\\hline &&\multicolumn{2}{l}{\textcircled{-2}}&&&\multicolumn{2}{l}{\textcircled{1}}&&\\ \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Diketahui grafik fungsi}\: \: f(x)=mx^{2}-2mx+m\: \: \textrm{berada di atas grafik fungsi}\\ &g(x)=2x^{2}-3,\: \textrm{maka nilai}\: \: m\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&m>2&&&\textrm{d}.&-6<m<2\\ \textrm{b}.&m>6&\textrm{c}.&2<m<6&\textrm{e}.&m<-6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}f(x)&=g(x)\\ mx^{2}-2mx+m&=2x^{2}-3\\ mx^{2}-2x^{2}-2mx+m+3&=0\\ \textrm{Supaya grafik}\: \: f(x)&\: \textrm{berada di atasnya, maka}\: \: D=B^{2}-4AC<0\\ (m-2)x^{2}-2mx+(m+3)&=0\begin{cases} A &=m-2 \\ B &=-2m \\ C &=m+3 \end{cases}\\ B^{2}-4AC&<0\\ (-2m)^{2}-4(m-2)(m+3)&<0\\ 4m^{2}-4\left ( m^{2}+m-6 \right )&<0\\ 4m^{2}-4m^{2}-4m+24&<0\\ -4m+24&<0\\ m-6&>0\\ m&>6 \end{aligned} \end{array}.

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Lanjutan BAB IV Sistem Pertidaksamaan Dua Variabel (K13 Revisi)

Sebelumnya kita buka arsip lama

A. Pertidaksamaan Linear Dua Variabel

\begin{array}{ll}\\ \textbf{\underline{Bentuk Umum}}&:\\ &\begin{cases} ax+by<c \\ ax+by\leq c \\ ax+by>c \\ ax+by\geq c \end{cases}\\\\ \textbf{Langkah-langkah}&\: \textrm{dalam membuat gambar grafik persamaan linear}\\ &\: \textrm{adalah sebagai berikut}:\\ &\bullet\quad \textrm{membuat gambar grafik}\: ax+by=c \: \textrm{untuk batas wilayahnya}\\ &\bullet \quad \textrm{menyelidiki wilayah yang dimaksud di sekitar garis}\: ax+by=c\\ &\bullet \quad \textrm{ambillah sebuah titik}\: \left ( x_{0},y_{0} \right )\: \textrm{sembarang, kemudian substitusikan}\\ &\: \: \quad \textrm{ke pertidaksamaan}\: \: ax+by\: ....\: c\\ &\bullet \quad \textrm{jika diperoleh nilai ketaksamaan yang benar, maka daerah di mana}\\ &\: \: \quad \textrm{titik uji}\: \left ( x_{0},y_{0} \right )\: \textrm{berada merupakan wilayah penyelesaiannya}\\ &\: \: \quad \textrm{demikian juga sebaliknya} \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Gambarlah himpunan penyelesaian (HP) dari pertidaksamaan linear berikut}\\ &\textrm{a}.\quad 3x+2y< 6\\ &\textrm{b}.\quad 3x+2y\leq 6\\ &\textrm{c}.\quad 3x+2y> 6\\ &\textrm{d}.\quad 3x+2y\geq 6\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Mula}&-\textrm{mula kita gambar garis}\: \: 3x+2y=6\\\\ &\qquad\qquad\qquad\begin{array}{|c|c|c|}\hline \textrm{Komponen}&\multicolumn{2}{|c|}{\textrm{Memotong}}\\\cline{2-3} \textrm{titik}&\textrm{sumbu}-y&\textrm{sumbu}-x\\\hline x&0&2\\\hline y&3&0\\\hline (x,y)&(0,3)&(2,0)\\\hline \end{array}\\\\ &\textrm{Selanjutnya gambar grafiknya sebagai berikut}. \end{aligned} \end{array}.

dan berikut untuk gambar wilayah sebagai himpunan penyelesaian  3x+2y<6.

Perhatikan posisi titik uji dan penjelasannya sesuai ilustrasi gambar di atas

\begin{array}{|c|c|c|c|}\hline \multicolumn{4}{|c|}{\textrm{Untuk wilayah disekitar garis}\: \: 3x+2y=6}\\\hline \textrm{No}&\textrm{Titik Uji}&\textrm{Pengujian}&\textrm{Keterangan untuk wilayah}\: \: 3x+2y<6\\\hline 1&(0,0)&3(0)+2(0)=0<\textbf{6}&\textrm{Di dalam wilayah}\\\hline 2&(0,1)&3(0)+2(1)=2<\textbf{6}&\textrm{Di dalam wilayah}\\\hline 3&(1,0)&3(1)+2(0)=3<\textbf{6}&\textrm{Di dalam wilayah}\\\hline 4&(1,1)&3(1)+2(1)=5<\textbf{6}&\textrm{Di dalam wilayah}\\\hline 5&(0,2)&3(0)+2(2)=4<\textbf{6}&\textrm{Di dalam wilayah}\\\hline 6&(2,0)&3(2)+2(0)=6=\textbf{6}&\textrm{Di luar wilayah}\\\hline 7&(2,2)&3(2)+2(2)=10>\textbf{6}&\textrm{Di luar wilayah}\\\hline 8&(0,3)&3(0)+2(3)=6=\textbf{6}&\textrm{Di luar wilayah}\\\hline 9&(3,0)&3(3)+2(0)=9>\textbf{6}&\textrm{Di luar wilayah}\\\hline 10&(3,3)&3(3)+2(3)=15>\textbf{6}&\textrm{Di luar wilayah}\\\hline \vdots &\vdots &\vdots&\vdots \\\hline \end{array}.

dan berikut juga gambar wilayah sebagai himpuanan penyelesaian dari  3x+2y\leq 6.

Anda boleh menyelidiki satu atau beberapa titik yang di sekitar garis  3x+2y=6, coba perhatikan tabel berikut:

\begin{array}{|c|c|c|c|c|}\hline \multicolumn{5}{|c|}{\textrm{Untuk wilayah disekitar garis}\: \: 3x+2y=6}\\\hline \textrm{No}&\textrm{Titik Uji}&\textrm{Pengujian}&\textrm{Keterangan untuk wilayah}\: \: 3x+2y\leq 6&\textrm{Kesimpulan}\\\hline 1&(0,0)&3(0)+2(0)=0<\textbf{6}&\textrm{Di dalam wilayah}&\textrm{Memenuhi}\\\hline 2&(0,1)&3(0)+2(1)=2<\textbf{6}&\textrm{Di dalam wilayah}&\textrm{Memenuhi}\\\hline 3&(1,0)&3(1)+2(0)=3<\textbf{6}&\textrm{Di dalam wilayah}&\textrm{Memenuhi}\\\hline 4&(1,1)&3(1)+2(1)=5<\textbf{6}&\textrm{Di dalam wilayah}&\textrm{Memenuhi}\\\hline 5&(0,2)&3(0)+2(2)=4<\textbf{6}&\textrm{Di dalam wilayah}&\textrm{Memenuhi}\\\hline 6&(2,0)&3(2)+2(0)=6=\textbf{6}&\textrm{Pada wilayah}&\textrm{Memenuhi}\\\hline 7&(2,2)&3(2)+2(2)=10>\textbf{6}&\textrm{Di luar wilayah}&\textrm{Tidak memenuhi}\\\hline 8&(0,3)&3(0)+2(3)=6=\textbf{6}&\textrm{Pada wilayah}&\textrm{Memenuhi}\\\hline 9&(3,0)&3(3)+2(0)=9>\textbf{6}&\textrm{Di luar wilayah}&\textrm{Tidak memenuhi}\\\hline 10&(3,3)&3(3)+2(3)=15>\textbf{6}&\textrm{Di luar wilayah}&\textrm{Tidak memenuhi}\\\hline \vdots &\vdots &\vdots&\vdots &\vdots \\\hline \end{array}.

B. Pertidaksamaan Kuadrat Dua Variabel

Proses menggambar grafik fungsi kuadrat anda bisa membuka arsip lama di sini

dan proses penentuan wilayah kurang lebih mirip seperti menentukan wilayah pada materi pertidaksamaan linear dua variabel di atas.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 2.&\textrm{Gambarlah himpunan penyelesaian (HP) dari pertidaksamaan kuadrat berikut}\\ &\textrm{a}.\quad y<x^{2}+2x-3\\ &\textrm{b}.\quad y\leq x^{2}+2x-3\\ &\textrm{c}.\quad y>x^{2}+2x-3\\ &\textrm{d}.\quad y\geq x^{2}+2x-3\\\\ &\textrm{Jawab}:\\ \end{array}.

Berikut wilayah untuk no. 2 a  sebagai himpunan penyelesaiannya

dan untuk wilayah no. 2 b  adalah

dan berikut juga untuk wilayah no. 2 d

C. Pertidaksamaan Kuadrat-Kuadrat Dua Variabel

Wilayah himpunan penyelesaiannya adalah tempat kedudukan titik-titik yang memenuhi kedua pertidaksamaan tersebut.

\begin{aligned} \textrm{Berikut contoh}&\, \: \textrm{pertidaksamaan kuadrat dua variabel}\\ y&<ax^{2}+bx+c\\ y&\leq ax^{2}+bx+c\\ y&>ax^{2}+bx+c\\ y&\geq ax^{2}+bx+c\\ \end{aligned}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 3.&\textrm{Gambarlah himpunan penyelesaian (HP) dari pertidaksamaan kuadrat-kuadrat berikut}\\\\ &\textrm{a}.\quad \begin{cases} y\geq x^{2}-4 \\ y\leq -x^{2}-x+2 \end{cases}\\\\ &\textrm{b}.\quad \begin{cases} y\leq 2x^{2}-9x+8 \\ y\geq -x^{2}-5x-4 \end{cases}\\\\ &\textrm{c}.\quad \begin{cases} x^{2}+y^{2}<16 \\ (x-2)^{2}+y^{2}>16 \end{cases}\\\\ &\textrm{d}.\quad\begin{cases} x^{2}-y^{2}\geq 5 \\ x^{2}+y^{2}\leq 13 \end{cases}\\\\ &\textrm{Jawab}:\\ \end{array}.

Untuk  no. 3 a , wilayah himpunan penyelesaiannya adalah irisan dari kedua pertidaksamaan kuadrat tersebut.

Berikut untuk wilayah pada no. 3 b ,

Dan untuk gambar  no. 3 c  adalah

Serta berikut untuk  no. 3 d.

bagian yang mendapat dobel aksiran baik  no.  3 c  maupun  no. 3  d   adalah merupakan daerah penyelesaian yang kita inginkan.

Sebagai catatannya untuk lebih memantapkan kebenaran gambar di atas cobalah pembaca yang budiman menguji dengan cara memasukkan beberapa titik uji.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Gambarlah himpunan penyelesaian (HP) dari pertidaksamaan linear berikut}\\ &\textrm{a}.\quad 4x+5y< 20\\ &\textrm{b}.\quad 4x+5y\leq 20\\ &\textrm{c}.\quad 4x+5y> 20\\ &\textrm{d}.\quad 4x+5y\geq 20\\\\ \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Gambarlah himpunan penyelesaian (HP) dari pertidaksamaan berikut}\\\\ &\textrm{a}.\quad \begin{cases} x+y<4 \\ y\geq x^{2}-9 \end{cases}\\\\ &\textrm{b}.\quad \begin{cases} 2y-x\geq 5 \\ x^{2}+y^{2}<25 \end{cases}\\\\ &\textrm{c}.\quad \begin{cases} y\leq -x^{2}+4 \\ y\leq x^{2}-2x-3 \end{cases}\\\\ &\textrm{d}.\quad\begin{cases} x^{2}+y^{2}\geq 25 \\ 2y-x\geq 5 \end{cases}\\\\  \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Gambarlah himpunan penyelesaian (HP) dari pertidaksamaan berikut}\\\\ &\textrm{a}.\quad \begin{cases} x^{2}+y\leq 4 \\ y\geq x^{2}-4 \end{cases}\\\\ &\textrm{b}.\quad \begin{cases} y< x^{2} \\ x> y^{2} \end{cases}\\\\ &\textrm{c}.\quad \begin{cases} y\geq x^{2}+6x \\ 4y> x^{2}-16 \end{cases}\\\\ &\textrm{d}.\quad\begin{cases} x^{2}+y^{2}\leq 25 \\ y<x^{2}-4x+5 \end{cases}\\\\ &\textrm{e}.\quad\begin{cases} x^{2}-y^{2}< 7 \\ x^{2}+y^{2}>25 \end{cases}\\\\ &\textrm{f}.\quad\begin{cases} x^{2}+y^{2}\geq 49 \\ (x-1)^{2}+y^{2}\leq 49 \end{cases}\\\\ \end{array}.

Sumber Referensi

  1. Kanginan, M,. 2017. Matematika untuk Siswa SMA-MA/SMA-MAK Kelas X (Cetakan ke-2). 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.

 

 

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BAB IV Sistem Pertidaksamaan Dua Variabel (K13 Revisi)

A. Sistem Persamaan Dua Variabel Linear Kuadrat (SPDVLK)

\begin{aligned}\textrm{Bentuk Umum}&\\ &\begin{cases} ax+by+c=0 \\ px^{2}+qy^{2}+rxy+sx+ty+u=0 \end{cases} \end{aligned}.

Perhatikanlah hal berikut ini,

\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Bentuk yang}&\: \textrm{sering muncul adalah}\\ &\begin{cases} y=ax+b \\ y=px^{2}+qx+r \end{cases}\\ & \end{aligned} }\\\hline \textrm{Penyelesaian}&\textrm{Proses}&\textrm{Keterangan}\\\hline y=y&\begin{aligned}ax+b&=px^{2}+qx+r\\ \textrm{di}&\textrm{ubah ke bentuk umum}\\ &Ax^{2}+Bx+C=0\\ &\textrm{dengan}\\ &D=B^{2}-4AC \end{aligned}&\begin{aligned} D&> 0\\ &\textrm{SPDVLK mempunyai 2 penyelesaian berbeda}\\ D&=0\\ &\textrm{SPDVLK mempunyai 1 penyelesaian saja}\\ D&< 0\\ &\textrm{SPDVLK tidak mempunyai penyelesaian} \end{aligned}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Carilah ada/tidaknya penyelesaian dari SPDVLK berikut ini}.\\\\ &\begin{cases} y=-x+2 \\ y=x^{2}-3x+2 \end{cases}\\\\ &\textrm{Jika ada, sketsalah tafsiran geometrinya}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui bahwa}:&\\ &\begin{cases} y=-x+2 \\ y= x^{2}-3x+2 \end{cases}\\ \textrm{maka}&\\ y&=y\\ x^{2}-3x+2&=-x+2\\ x^{2}-2x&=0\\ x(x-2)&=0\\ x=0\quad \textrm{V}\quad x&=2\\ \textrm{selanjutnya}&\: \textbf{hasil ini dapat kita substitusikan ke}\: \: \: \: y=-x+2\\ &\textrm{untuk mendapatkan nilai}\: \: \: \: y\: \: \: \: \textrm{yaitu}:\\ &\begin{cases} x=0 & \Rightarrow y=-0+2=2 ,\quad \textrm{diperoleh titik}\quad (0,2)\\ x=2 & \Rightarrow y=-2+2=0 ,\quad \textrm{diperoleh juga titik}\quad (2,0) \end{cases}\\ &\\ \textrm{sehingga HP-nya}&=\left \{ (0,2),(2,0) \right \}\end{aligned} \end{array}.

Berikut Sketsa grafiknya

\begin{array}{ll}\\ 2.&\textrm{Carilah ada/tidaknya penyelesaian dari SPDVLK berikut ini}.\\\\ &\begin{cases} 2x+3y-1=0 \\ x^{2}-2y^{2}-xy-x-4y-2=0 \end{cases}\\\\ &\textrm{Jika ada, sketsalah tafsiran geometrinya}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui bahwa}:&\\ &\begin{cases} 2x+3y-1=0 &\Rightarrow \quad y=-\displaystyle \frac{2}{3}x+\frac{1}{3}\\ x^{2}-2y^{2}-xy-x-4y-2=0 \end{cases}\\ \textrm{maka}&\: \textrm{substitusikan}\: \: \: y=-\displaystyle \frac{2}{3}x+\frac{1}{3}\quad \textrm{ke}\quad x^{2}-2y^{2}-xy-x-4y-2=0\\ x^{2}-2\left ( -\displaystyle \frac{2}{3}x+\frac{1}{3} \right )^{2}&-x\left ( -\displaystyle \frac{2}{3}x+\frac{1}{3} \right )-x-4\left ( -\displaystyle \frac{2}{3}x+\frac{1}{3} \right )-2=0\\ x^{2}-2\left ( \displaystyle \frac{4}{9}x^{2}-\frac{4}{9}x+\frac{1}{9} \right )&+\frac{2}{3}x^{2}-\frac{1}{3}x-x+\frac{8}{3}x-\frac{4}{3}-2=0\\ x^{2}-\displaystyle \frac{8}{9}x^{2}+\frac{8}{9}x-\frac{2}{9}&+\frac{2}{3}x^{2}-\frac{1}{3}x-x+\frac{8}{3}x-\frac{4}{3}-2=0\\ \displaystyle \frac{7}{9}x^{2}+\frac{20}{9}x-\frac{32}{9}&=0\\ 7x^{2}+20x-32&=0\\ (x+4)(7x-8)&=0\\  \textrm{selanjutnya}&\: \textbf{hasil ini dapat kita substitusikan ke}\: \: \: \: y=-\displaystyle \frac{2}{3}x+\frac{1}{3}\\ &\textrm{untuk mendapatkan nilai}\: \: \: \: y\: \: \: \: \textrm{yaitu}:\\ &\begin{cases} x=-4 & \Rightarrow y=3 ,\quad \textrm{diperoleh titik}\quad (-4,3)\\ x=\displaystyle \frac{8}{7} & \Rightarrow y=-\displaystyle \frac{3}{7} ,\quad \textrm{diperoleh juga titik}\quad \left ( \displaystyle \frac{8}{7},-\frac{3}{7} \right ) \end{cases}\\ &\\ \textrm{sehingga HP-nya}&=\left \{ (-4,3),\left ( \displaystyle \frac{8}{7},-\frac{3}{7} \right ) \right \}\end{aligned} \end{array}.

Dan berikut sketsa grafiknya

\begin{array}{ll}\\ 3.&\textrm{Tunjukkanlah bahwa sistem persamaan berikut}:\\ &\begin{cases} ab=1 \\ bc=2 \\ cd=3 \\ de=4 \\ ea=6 \end{cases}\\ &\textrm{memiliki penyelesaian}\: \: \left ( -\displaystyle \frac{3}{2},-\frac{2}{3},-3,-1,-4 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{3}{2},\frac{2}{3},3,1,4 \right )\\\\ &\textbf{Bukti}:\\ &\begin{aligned}ab\times bc\times cd\times de\times ea&=1\times 2\times 3\times 4\times 6\\ \left ( abcde \right )^{2}&=144\\ abcde&=\pm \sqrt{144}\\ abcde&=\pm 12\\ (ab).c.(de)&=\pm 12\\ 1.c.4&=\pm 12\\ c&=\displaystyle \frac{\pm 12}{4}\\ &=\pm 3\\ \textrm{dari sini kita}&\: \textrm{akan mendapatkan pula nilai yang lainnya, antara lain}:\\ bc&=2\\ b&=\displaystyle \frac{2}{c}\\ b&=\displaystyle \frac{2}{\pm 3}\\ b&=\pm \displaystyle \frac{2}{3}\\ \textrm{nilai yang}&\: \textrm{lainpun akan ketemu juga, yaitu}\\ a&=\pm \displaystyle \frac{3}{2}\\ d&=\pm 1\\ e&=\pm 4\qquad\quad \blacksquare \end{aligned} \end{array}.

B. Sistem Persamaan Dua Variabel Kuadrat Kuadrat (SPDVKK)

\begin{aligned}\textrm{Bentuk Umum}&\\ &\begin{cases} ax^{2}+by^{2}+cxy+dx+ey+f=0 \\ px^{2}+qy^{2}+rxy+sx+ty+u=0 \end{cases} \end{aligned}.

\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Bentuk yang}&\: \textrm{sering muncul adalah}\\ &\begin{cases} y=ax^{2}+bx+c \\ y=px^{2}+qx+r \end{cases}\\ & \end{aligned} }\\\hline \textrm{Penyelesaian}&\textrm{Proses}&\textrm{Keterangan}\\\hline y=y&\begin{aligned}ax^{2}+bx+c&=px^{2}+qx+r\\ \textrm{dengan}&\: \textrm{syarat}\\ a-p&\neq 0\\ &\textrm{diubah kebentuk umum}\\ &Ax^{2}+Bx+C=0\\ &\textrm{dengan}\\ &D=B^{2}-4AC \end{aligned}&\begin{aligned} D&> 0\\ &\textrm{SPDVKK mempunyai 2 penyelesaian berbeda}\\ D&=0\\ &\textrm{SPDVKK mempunyai 1 penyelesaian saja}\\ D&< 0\\ &\textrm{SPDVKK tidak mempunyai penyelesaian} \end{aligned}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Carilah ada/tidaknya penyelesaian dari SPDVKK berikut ini}.\\\\ &\begin{cases} y=-x^{2} \\ y=x^{2}+2x+1 \end{cases}\\\\ &\textrm{dan sketsalah tafsiran geometrinya}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui bahwa}:&\\ &\begin{cases} y=-x^{2}\\ y=x^{2}+2x+1 \end{cases}\\ &\: \textrm{substitusikan}\: \: \: y=-x^{2}\quad \textrm{ke}\quad y=x^{2}+2x+1\\ \textrm{sehingga diperoleh}&\\ -x^{2}&=x^{2}+2x+1\\ 2x^{2}+2x+1&=0\: \begin{cases} a=2 \\ b=-2 \\ c=1 \end{cases}\\ \textrm{Karena, nilai}&\: \: D=b^{2}-4ac=(-2)^{2}-4(2)(1)=4-8=-4< 0\\ \textrm{maka diperoleh}&\: \textrm{keterangan bahwa kedua parabola tersebut}\\ &\textbf{tidak berpotongan dan tidak bersinggungan}. \end{aligned} \end{array}.

Berikut sketsa grafiknya

\begin{array}{ll}\\ 2.&\textrm{Carilah nilai untuk}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{dari SPDVKK berikut ini}.\\\\ &\begin{cases} x^{2}+xy=12 \\ 2xy+3y^{2}=-5 \end{cases}\\\\ &\textrm{Jawab}:\\ &\begin{array}{lllllllllllll}\\ x^{2}+xy&=12&\times (5)&5x^{2}+5xy&=60\\ 2xy+3y^{2}&=-5&\times (12)&24xy+36y^{2}&=-60&+\\\cline{1-5} &&&5x^{2}+29xy+36y^{2}&=0\\ &&&(x+4y)(5x+9y)&=0\\ &&&x=-4y\quad \textbf{V}\quad x&=-\displaystyle \frac{9}{5}\\\\ &&&\underline{\textbf{untuk}}\\ &&&\bullet \quad x=-4y\quad\: \Rightarrow &(x,y)&=\begin{cases} (-4,1) \\ (4,-1) \end{cases}\\\\ &&&\underline{\textbf{dan untuk}}\\ &&&\bullet \quad x=-\displaystyle \frac{9}{5}y\quad\: \Rightarrow &(x,y)&=\begin{cases} \left (-3\sqrt{3},\displaystyle \frac{5}{3}\sqrt{3} \right ) \\\\ \left (3\sqrt{3},-\displaystyle \frac{5}{3}\sqrt{3} \right ) \end{cases} \end{array} \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Diberikan 3 persamaan berikut dengan}\: \: x,y,z> 0\\ &\begin{cases} (x-1)(y-2)=12 \\ (y-2)(z-3)=20 \\ (z-3)(x-1)=15 \end{cases}\\ &\textrm{Tentukanlah nilai dari}\: \: 3x+2y+3z\\\\ &\textrm{Jawab}:\\ &\begin{aligned}(x-1)(y-2)\times (y-2)(z-3)&\times (z-3)(x-1)=12\times 20\times 15\\ \left ((x-1)(y-2)(z-3) \right )^{2}&=3600\\ (x-1)(y-2)(z-3)&=\pm \sqrt{3600}\\ &=\pm 60\\ \left ((x-1)(y-2) \right )(z-3)&=\pm 60\\ 12(z-3)&=\pm 60\\ (z-3)&=\displaystyle \frac{\pm 60}{12}\\ (z-3)&=\pm 5\\ z&=\pm 5+3\qquad (\textrm{karena \textit{z} positif, maka})\\ z&=5+3\\ &=8\\ \textrm{untuk nilai yang}&\: \textrm{lain pencariannya diserahkan kepada pembaca yang budiman} \end{aligned} \end{array}.

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Tentukanlah HP dari SPDVLK berikut dan sketsalah grafiknya}\\ &\begin{array}{llllll}\\ \textrm{a}.&\begin{cases} y=2x+8 \\ y=x^{2} \end{cases}&\textrm{f}.&\begin{cases} y+x-3=0 \\ y=x^{2}-4x+3 \end{cases}\\\\ \textrm{b}.&\begin{cases} y=x^{2}-1 \\ y=3x+1 \end{cases}&\textrm{g}.&\begin{cases} 4x-y=5 \\ 2x^{2}-3xy-2y^{2}+7x+6y-4=0 \end{cases}\\\\ \textrm{c}.&\begin{cases} y=x \\ y=3x-x^{2} \end{cases}&\textrm{h}.&\begin{cases} x-y=-1 \\ 2x^{2}-xy-y^{2}+2x-2y+6=0 \end{cases}\\\\ \textrm{d}.&\begin{cases} y=x-2 \\ y=x^{2}-5x+4 \end{cases}&\textrm{i}.&\begin{cases} 2x-y=3 \\ x^{2}+xy+y+1=0 \end{cases}\\\\ \textrm{e}.&\begin{cases} y=2x-1 \\ y=x^{2}-1 \end{cases}&\textrm{j}.&\begin{cases} 2x-3y+4=0 \\ x^{2}-25y^{2}-4x-30y-5=0 \end{cases}\\ \end{array}\\ \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah HP dari SPDVKK berikut}\\ &\begin{array}{llllll}\\ \textrm{a}.&\begin{cases} y=4-x^{2} \\ y=x^{2}+1 \end{cases}&\textrm{f}.&\begin{cases} x^{2}+y^{2}=5 \\ x^{2}+3xy+y^{2}=11 \end{cases}\\\\ \textrm{b}.&\begin{cases} y=9-x^{2} \\ y=-2x^{2}-4x+1 \end{cases}&\textrm{g}.&\begin{cases} y=x^{2}+1 \\ y=2x^{2}-3x+3 \end{cases}\\\\ \textrm{c}.&\begin{cases} y=x^{2}-3x \\ y=x^{2}-2x+6 \end{cases}&\textrm{h}.&\begin{cases} x^{2}-y^{2}=25 \\ 3y^{2}-16x=0 \end{cases}\\\\ \textrm{d}.&\begin{cases} y=-x^{2}+2x+3 \\ y=x^{2}-4x+6 \end{cases}&\textrm{i}.&\begin{cases} x^{2}+y^{2}=74 \\ xy=37 \end{cases}\\\\ \textrm{e}.&\begin{cases} y=x^{2}+x \\ y=2x^{2}-3x \end{cases}&\textrm{j}.&\begin{cases} y=x^{2}-4 \\ y=-x^{2}+x+4 \end{cases}\\ \end{array}\\ \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Tentukanlah penyelesaian untuk sistem persamaan berikut}\\ &\begin{cases} x+y+z=6 \\ x^{2}+y^{2}+z^{2}=14 \\ xyz=6 \end{cases}\\ \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah penyelesaian untuk sistem persamaan berikut}\\ &\begin{cases} x+y+xy=11 \\ y+z+yz=14 \\ z+x+zx=19 \end{cases}\\ \end{array}.

Sumber Referensi

  1. Tung, K. Y., 2012. Pintar Matematika SMA Kelas X untuk olimpiade dan Pengayaan pelajaran. Yogyakarta: ANDI.
  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.

 

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BAB III Sistem Persamaan Linear Tiga Variabel (K13 Revisi)

B. Sistem Persamaan Linear Tigaa Variabel (SPLTV)

Sebelumnya silahkan merujuk ke

\begin{array}{ll}\\ \underline{\textbf{Bentuk Umum}}&:\\ &\begin{cases} a_{1}x+b_{1}y+c_{1}z=d_{1} \\ a_{2}x+b_{2}y+c_{2}z=d_{2} \\ a_{3}x+b_{3}y+c_{3}z=d_{3} \end{cases}\\\\ \qquad \quad \textbf{dengan}&\bullet \quad a_{1},\: a_{2},\: a_{3},\\ &\, \: \: \quad b_{1},\: b_{2},\: b_{3},\\ &\, \: \: \quad c_{1},\: c_{2},\: c_{3},\\ &\, \: \: \quad d_{1},\: d_{2},\: d_{3}\\ &\: \: \quad \textrm{semuanya adalah bilangan real} \end{array}.

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