Lanjutan: Bilangan Bentuk Akar dan Pangkat Pecahan

Sebelumnya bisa Anda menuju di sini

\begin{array}{|c|l|l|c|}\hline \textbf{No}&\textbf{\: \: \qquad Bentuk Akar}&\textbf{\: \: \: \: Pangkat pecahan}\\\hline 1&\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}&a^{n}=b\quad \Rightarrow \quad a=b^{^{^{\frac{1}{n}}}}=\sqrt[n]{b}\\\hline 2&\sqrt{\displaystyle \frac{a}{b}}=\displaystyle \frac{\sqrt{a}}{\sqrt{b}}&a> 0\quad \Rightarrow \quad \sqrt[n]{a}\geq 0\\\hline 3&a\sqrt{c}\pm b\sqrt{c}=\left ( a\pm b \right )\sqrt{c}&a< 0\begin{cases} n & \text{ genap } \Rightarrow \sqrt[n]{a}< 0\\ n & \text{ ganjil } \Rightarrow \sqrt[n]{a}\quad \textbf{bukan riil} \end{cases}\\\hline 4&\sqrt{\left ( a+b \right )\pm 2\sqrt{ab}}=\left (\sqrt{a} \pm \sqrt{b}\right )&a^{^{^{^{\frac{\begin{matrix} \\ m \end{matrix}}{\begin{matrix} n\\ \end{matrix}}}}}}=\sqrt[n]{a^{m}}\\\hline 5&\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a}{b}\sqrt{b}&\\\cline{1-2} 6&\displaystyle \frac{c}{a\pm \sqrt{b}}=\frac{c}{a\pm \sqrt{b}}\times \frac{a\mp \sqrt{b}}{a\mp \sqrt{b}}=\frac{c\left ( a\mp \sqrt{b} \right )}{a^{2}- b}&\\\cline{1-2} 7&\begin{aligned}\displaystyle \frac{c}{\sqrt{a}\pm \sqrt{b}}&=\frac{c}{\sqrt{a}\pm \sqrt{b}}\times \frac{\sqrt{a}\mp \sqrt{b}}{\sqrt{a}\mp \sqrt{b}}\\ &=\frac{c\left ( \sqrt{a}\mp \sqrt{b} \right )}{a- b} \end{aligned}&\\\cline{1-2} 8&\displaystyle \frac{a}{\sqrt[n]{b^{m}}}=\frac{a}{\sqrt[n]{b^{m}}}\times \frac{\sqrt[n]{b^{n-m}}}{\sqrt[n]{b^{n-m}}}=\frac{a}{b}\sqrt[n]{b^{n-m}}&\\\cline{1-2} 9&\begin{aligned}\displaystyle \frac{c}{\sqrt[3]{a}\pm \sqrt[3]{b}}=&\frac{c}{\sqrt[3]{a}\pm \sqrt[3]{b}}\\ &\times \frac{\sqrt[3]{a^{2}}\mp \sqrt[3]{ab}+\sqrt[3]{b^{2}}}{\sqrt[3]{a^{2}}\mp \sqrt[3]{ab}+\sqrt[3]{b^{2}}} \end{aligned}&\\\hline \end{array}.

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

\begin{array}{ll}\\ \fbox{1}.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{3^{\frac{5}{6}}\times 12^{\frac{7}{12}}}{6^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\: \: \textrm{adalah} ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad 6^{\frac{1}{4}}&&\textrm{d}.\quad \left ( \displaystyle \frac{2}{3} \right )^{\frac{3}{4}}\\ \textrm{b}.\quad 6^{\frac{3}{4}}&\textrm{c}.\quad 6^{\frac{2}{3}}&\textrm{e}.\quad \left ( \displaystyle \frac{3}{2} \right )^{\frac{3}{4}}\end{array}\\\\ &\textrm{Jawab}:\qquad \textbf{b}\\ &\begin{aligned}\displaystyle \frac{3^{\frac{5}{6}}\times 12^{\frac{7}{12}}}{6^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}&=\displaystyle \frac{3^{\frac{5}{6}}\times (3\times 4)^{\frac{7}{12}}}{(2\times 3)^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 4^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times \left ( 2^{2} \right )^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 2^{\frac{14}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=2^{(\frac{14}{12}-\frac{2}{3}+\frac{1}{4})}\times 3^{(\frac{5}{6}+\frac{7}{12}-\frac{2}{3})}\\ &=2^{(\frac{14-8+3}{12})}\times 3^{(\frac{10+7-8}{12})}\\ &=2^{\frac{9}{12}}\times 3^{\frac{9}{12}}\\ &=(2\times 3)^{\frac{9}{12}}\\ &=6^{\frac{9}{12}}\\ &=6^{\frac{3}{4}} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Bentuk} \: \: \left ( p^{-1}+q^{-1} \right )^{-1}\: \: \textrm{akan senilai dengan}...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad p+q&&\textrm{d}.\quad \displaystyle \frac{1}{pq}\\\\ \textrm{b}.\quad \displaystyle \frac{pq}{p+q}\quad &\textrm{c}.\quad p\quad &\textrm{e}.\quad \displaystyle \frac{p+q}{pq} \end{array}\\\\\\ &\textrm{Jawab}:\qquad \textbf{b}\\ &\begin{aligned}\left ( p^{-1}+q^{-1} \right )^{-1}&=\displaystyle \frac{1}{\left ( p^{-1}+q^{-1} \right )}\\ &=\displaystyle \frac{1}{\displaystyle \frac{1}{p}+\frac{1}{q}}\\ &=\displaystyle \frac{1}{\displaystyle \frac{1}{p}+\frac{1}{q}}\times \displaystyle \frac{pq}{pq}\\ &=\displaystyle \frac{pq}{q+p},\qquad \textnormal{atau}\\ &=\displaystyle \frac{pq}{p+q} \end{aligned} \end{array}.

Dalam penyelesaian soal seperti 2 nomor di atas tidak harus seperti itu, apa yang bisa Anda lakukan mungkin lebih ringkas, karena contoh di atas hanyalah sebagai salah satu alternatif saja.

Coba perhatikan penyelesaian soal berikut!

\begin{array}{ll}\\ \fbox{3}.&\textrm{Penyelesaian untuk persamaan} \: \: 2^{5}\times 8^{\frac{2}{3}}=\left ( \displaystyle \frac{1}{2} \right )^{3-2x}\: \: \textrm{adalah}...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2&&\textrm{d}.\quad \displaystyle 6\\\\ \textrm{b}.\quad \displaystyle 3\quad &\textrm{c}.\quad 5\quad &\textrm{e}.\quad \displaystyle 8 \end{array}\\\\\\ &\textrm{Jawab}:\qquad \textbf{c}\\ &\textrm{Diketahui bahwa}:\\ &\qquad \begin{array}{|c|c|} \multicolumn{2}{c}{\begin{aligned}&\\ &2^{5}\times 8^{\frac{2}{3}}=\left ( \displaystyle \frac{1}{2} \right )^{3-2x}\\ & \end{aligned}}\\\hline \begin{aligned}&\\ 2^{5}\times 8^{\frac{2}{3}}&=\left ( \displaystyle \frac{1}{2} \right )^{3-2x}\\ 2^{5}\times \left ( 2^{3} \right )^{\frac{2}{3}}&=\left ( 2^{-1} \right )^{3-2x}\\ 2^{5}\times 2^{2}&=2^{-3+2x}\\ 2^{5+2}&=2^{2x-3}\\ \textrm{kita}&\: \textrm{ambil}\\ 5+2&=2x-3\\ 7&=2x-3\\ 10&=2x\\ 2x&=10\\ x&=5\\ & \end{aligned}&\begin{aligned}&\\ 2^{5}\times 8^{\frac{2}{3}}&=\left ( \displaystyle \frac{1}{2} \right )^{3-2x}\\ 2^{5}\times \left ( 2^{3} \right )^{\frac{2}{3}}&=\displaystyle \frac{1}{2^{3-2x}}\\ 2^{5}\times 2^{2}\times 2^{3-2x}&=1\\ 2^{5+2+3-2x}&=1\\ 2^{10-2x}&=1\\ 2^{10-2x}&=2^{0}\\ \textrm{semisal dengan}&\: \textrm{cara sebelah kiri}\\ 10-2x&=0\\ -2x&=-10\\ x&=5\\ & \end{aligned}\\\hline \end{array} \end{array}.

Mungkin di antara Anda akan bertanya mengapa 1 dapat digantikan dengan  2^{0}  ?

\begin{array}{ll}\\ \fbox{4}.&\textrm{Harga \textit{x} untuk persamaan}\\ &\left ( \displaystyle \frac{1}{\sqrt{2}} \right )^{^{^{\begin{matrix} 2x-6\\ \end{matrix}}}}=\sqrt{2^{^{^{\begin{matrix} 4x+2\\ \end{matrix}}}}}\\ &\begin{array}{llllllll}\\ \textrm{a}.&\displaystyle \frac{2}{3}&&&\textrm{d}.&2\\\\ \textrm{b}.&\displaystyle \frac{4}{3}&\textrm{c}.&\displaystyle \frac{5}{3}&\textrm{e}.&3 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\left ( \displaystyle \frac{1}{\sqrt{2}} \right )^{^{^{\begin{matrix} 2x-6\\ \end{matrix}}}}&=\sqrt{2^{^{^{\begin{matrix} 4x+2\\ \end{matrix}}}}}\\ \left ( 2^-{^{\frac{1}{2}}} \right )^{^{^{\begin{matrix} 2x-6\\ \end{matrix}}}}&=2^{^{^{\begin{matrix} \frac{4x+2}{2}\\ \end{matrix}}}}\\ 2^{-x+3}&=2^{2x+1}\\ -x+3&=2x+1\\ -3x&=-2\\ x&=\displaystyle \frac{2}{3} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Perhatikanlah hubungan berikut}\: \: 2^{p}=3^{q}=6^{r}.\\ &\textrm{Tunjukkanlah bahwa}\: \: pr+qr-pq=0\\\\ &\textbf{Bukti}\\ &\begin{aligned}2^{p}&=3^{q}=6^{r}\begin{cases} 2^{p} &=3^{q} \quad \Rightarrow \quad 2=3^{^{^{^{\frac{q}{p}}}}} \quad \textrm{atau}\quad 2^{^{^{^{\frac{p}{q}}}}}=3\: ............\textcircled{1}\\ 2^{p} &=6^{r} \quad \Rightarrow \quad 2=6^{^{^{^{\frac{r}{p}}}}}\: .................................\textcircled{2}\\ 3^{q} &=6^{r} \quad \Rightarrow \quad 3=6^{^{^{^{\frac{r}{q}}}}}\: ..............................\textcircled{3} \end{cases}\\ 2^{p}&=6^{r}\\ 2^{p}&=(2\times 3)^{r}\\ 2^{p}&=2^{r}\times 3^{r}\\ 2^{p}&=2^{r}\times \left ( 2^{^{^{^{\frac{p}{q}}}}} \right )^{r}\\ 2^{p}&=2^{r}\times 2^{^{^{^{\frac{pr}{q}}}}}\\ 2^{p}&=2^{^{^{^{r+\frac{pr}{q}}}}}\\ p&=r+\frac{pr}{q}\\ pq&=qr+pr\\ 0&=pr+qr-pq\qquad \textrm{atau}\\ pr+qr-pq&=0\qquad\quad \blacksquare \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Harga}\: \: a+b\: \: \textrm{jika}\: \: \sqrt{\displaystyle \frac{8+4\sqrt{3}}{8-4\sqrt{3}}}=a+b\sqrt{3}\: \: \textrm{adalah}...\: .\\ &\begin{array}{llllllll}\\ \textrm{a}.&\displaystyle 1&&&\textrm{d}.&4\\\\ \textrm{b}.&\displaystyle 2&\textrm{c}.&\displaystyle 3&\textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\sqrt{\displaystyle \frac{8+4\sqrt{3}}{8-4\sqrt{3}}}&=a+b\sqrt{3}\\ \sqrt{\displaystyle \frac{8+2.2.\sqrt{3}}{8-2.2.\sqrt{3}}}&=a+b\sqrt{3}\\ \sqrt{\displaystyle \frac{8+2\sqrt{4}\sqrt{3}}{8-2\sqrt{4}\sqrt{3}}}&=a+b\sqrt{3}\\ \sqrt{\displaystyle \frac{8+2\sqrt{12}}{8-2\sqrt{12}}}&=a+b\sqrt{3}\\ \displaystyle \frac{\sqrt{6+2+2\sqrt{6.2}}}{\sqrt{6+2-2\sqrt{6.2}}}&=a+b\sqrt{3}\\ \displaystyle \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}&=a+b\sqrt{3}\\ \displaystyle \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}&=a+b\sqrt{3}\\ \displaystyle \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\times \displaystyle \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}&=a+b\sqrt{3}\\ \displaystyle \frac{6+2\sqrt{6}\sqrt{2}+2}{6-2}&=a+b\sqrt{3}\\ \displaystyle \frac{8+2.2\sqrt{3}}{4}&=a+b\sqrt{3}\\ 2+1.\sqrt{3}&=a+b\sqrt{3}\\ \textrm{sehingga}&\\ &\begin{cases} &\begin{cases} a & = 2\\ b & =1 \end{cases} \\\\ &\textrm{maka}\\ & a+b=3 \end{cases} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt{3,6}+\sqrt{0,4}}{\sqrt{10}-\sqrt{3,6}-\sqrt{0,4}}\: \: \textrm{adalah}...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad 4\\ \textrm{b}.\quad 2\quad &\textrm{c}.\quad 3\quad &\textrm{e}.\quad 5\end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\displaystyle \frac{\sqrt{3,6}+\sqrt{0,4}}{\sqrt{10}-\sqrt{3,6}-\sqrt{0,4}}&=\displaystyle \frac{\sqrt{\displaystyle \frac{36}{10}}+\sqrt{\displaystyle \frac{4}{10}}}{\sqrt{\displaystyle \frac{100}{10}}-\sqrt{\displaystyle \frac{36}{10}}-\sqrt{\displaystyle \frac{4}{10}}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{\sqrt{10}}\left ( \sqrt{36}+\sqrt{4} \right )}{\displaystyle \frac{1}{\sqrt{10}}\left ( \sqrt{100}-\sqrt{36}-\sqrt{4} \right )}\\ &=\displaystyle \frac{(6+2)}{(10-6-2)}\\ &=\displaystyle \frac{8}{2}\\ &=4 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Jika}\: \: \displaystyle p+\sqrt{p}=1\: \: \textrm{maka}\: \: p+\displaystyle \frac{1}{p}=...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad 0&&\textrm{d}.\quad 3\\ \textrm{b}.\quad 1\quad &\textrm{c}.\quad 2\quad &\textrm{e}.\quad 4\end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}:\\ &\begin{aligned}p+\sqrt{p}&=1\\ \left ( \sqrt{p} \right )^{2}+\sqrt{p}&=1\: ,\qquad \textrm{di bagi dengan}\: \: \: \sqrt{p}\\ \displaystyle \frac{\left ( \sqrt{p} \right )^{2}}{\sqrt{p}}+\frac{\sqrt{p}}{\sqrt{p}}&=\displaystyle \frac{1}{\sqrt{p}}\\ \sqrt{p}+1&=\displaystyle \frac{1}{\sqrt{p}}\\ \sqrt{p}-\displaystyle \frac{1}{\sqrt{p}}&=-1\\ \left ( \sqrt{p}-\displaystyle \frac{1}{\sqrt{p}} \right )^{2}&=(-1)^{2}\\ p-2+\displaystyle \frac{1}{p}&=1\\ p+\displaystyle \frac{1}{p}&=3 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Jika}\: \: \displaystyle q+\displaystyle \frac{1}{q}=3\: \: \textrm{maka}\: \: q^{3}+\displaystyle \frac{1}{q^{3}}=...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad 12&&\textrm{d}.\quad 21\\ \textrm{b}.\quad 15\quad &\textrm{c}.\quad 18\quad &\textrm{e}.\quad 24\end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}q+\displaystyle \frac{1}{q}=3\quad \Leftrightarrow \quad &\left ( q+\displaystyle \frac{1}{q} \right )^{3}=3^{3}\\ &q^{3}+3.q^{2}.\displaystyle \frac{1}{q}+3.q.\displaystyle \frac{1}{q^{2}}+\displaystyle \frac{1}{q^{3}}&=27\\ &q^{3}+3q+\displaystyle \frac{3}{q}+\frac{1}{q^{3}}=27\\ &q^{3}+3\left ( q+\displaystyle \frac{1}{q} \right )+\displaystyle \frac{1}{q^{3}}=27\\ &q^{3}+3.3+\displaystyle \frac{1}{q^{3}}=27\\ &q^{3}+\displaystyle \frac{1}{q^{3}}+9=27\\ &q^{3}+\displaystyle \frac{1}{q^{3}}=27-9\\ &q^{3}+\displaystyle \frac{1}{q^{3}}=18 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Nilai dari}\: \: \displaystyle \sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\: \: \: \textrm{adalah}...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad \sqrt[3]{10}\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\quad &\textrm{c}.\quad \sqrt[3]{2}\quad &\textrm{e}.\quad \sqrt[3]{65}\end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\left ( 1\pm \sqrt{13} \right )^{3}&=1^{3}\pm 3.1^{2}.\sqrt{13}+3.1.\left ( \sqrt{13} \right )^{2}\pm \left ( \sqrt{13} \right )^{3}\\ &\begin{cases} \left ( 1+\sqrt{13} \right )^{3} & =1+3\sqrt{13}+39+13\sqrt{13}=40+16\sqrt{13}=8\left ( 5+2\sqrt{13} \right ) \\ \left ( 1-\sqrt{13} \right )^{3} & =1-3\sqrt{13}+39-13\sqrt{13}=40-16\sqrt{13}=8\left ( 5-2\sqrt{13} \right ) \end{cases}\\ &\textrm{maka} \\ &\begin{cases} 8\left ( 5+2\sqrt{13} \right ) & =\left ( 1+\sqrt{13} \right )^{3}\Rightarrow \left ( 5+2\sqrt{13} \right )=\displaystyle \frac{\left ( 1+\sqrt{13} \right )^{3}}{8}=\left ( \displaystyle \frac{1+\sqrt{13}}{2} \right )^{3} \\ 8\left ( 5+2\sqrt{13} \right ) & =\left ( 1-\sqrt{13} \right )^{3}\Rightarrow \left ( 5-2\sqrt{13} \right )=\displaystyle \frac{\left ( 1-\sqrt{13} \right )^{3}}{8}=\left ( \displaystyle \frac{1-\sqrt{13}}{2} \right )^{3} \end{cases}\\ &\textrm{Sehingga}\\ &\begin{cases} \left ( \displaystyle \frac{1+\sqrt{13}}{2} \right )^{3} & =5+2\sqrt{13}\: \Rightarrow \left ( \displaystyle \frac{1+\sqrt{13}}{2} \right )=\displaystyle \sqrt[3]{5+2\sqrt{13}}\\ \left ( \displaystyle \frac{1-\sqrt{13}}{2} \right )^{3} & =5-2\sqrt{13}\: \Rightarrow \left ( \displaystyle \frac{1-\sqrt{13}}{2} \right )=\displaystyle \sqrt[3]{5-2\sqrt{13}} \end{cases}\\ \textrm{Jadi}&,\\ \displaystyle \sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}&=\left ( \displaystyle \frac{1+\sqrt{13}}{2} \right )+\left ( \displaystyle \frac{1-\sqrt{13}}{2} \right )\\ &=\displaystyle \frac{2}{2}\\ &=1 \end{aligned} \end{array}.

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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