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May 26, 2018 Bank Soal, Bentuk Akar, Eksponen, SNMPTN dan SBMPTN

Matematika Dasar Bentuk Akar (*Soal Dari Aneka Macam Sumber)

Kesulitan menganalisa kalimat soal

$\sqrt{a}=a^{\frac{1}{2}}$

$\sqrt[n]{a^{m}}=a^{\frac{m}{n}}$

$\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}$

$\left (\sqrt{a}+\sqrt{b} \right )\left (\sqrt{a}-\sqrt{b} \right )=a-b$

$\left (a+\sqrt{b} \right )\left (a-\sqrt{b} \right )=a^{2}-b$

$\left (\sqrt{a}+b \right )\left (\sqrt{a}-b \right )=a-b^{2}$

$\frac{c}{\sqrt{a}+\sqrt{b}}=\frac{c\left (\sqrt{a}-\sqrt{b}\right)}{a-b}$

$\sqrt{(a+b)+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}$ dengan $a,\ b \geq 0$

$\sqrt{(a+b)-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}$ dengan $a,\ b \geq 0$ dan $a \geq b$

$\sqrt{a\sqrt{a\sqrt{a\sqrt{\cdots}}}}=a$ dengan $a \geq 0$

$\sqrt{a \cdot b +\sqrt{ a \cdot b +\sqrt{a \cdot b +\sqrt{\cdots}}}}=a$ dengan $a-b=1$

$\sqrt{a \cdot b -\sqrt{ a \cdot b -\sqrt{a \cdot b -\sqrt{\cdots}}}}=b$ dengan $a-b=1$

$\left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2}$

$\left ( a+b \right )^{2}=a^{2}+2ab+b^{2}$

$\left ( a-b \right )^{2}=a^{2}-2ab+b^{2}$

1. Soal SPMB 2006 Kode 510Jika bilangan lingkaran $a$ dan $b$ memenuhi $\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30}$ maka $ab=\cdots$
$(A)\ -22$
$(B)\ -11$
$(C)\ -9$
$(D)\ 2$
$(E)\ 13$Alternatif Pembahasan:

Hint

$\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30}$
$\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}} \cdot \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}$
$\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=\frac{5+6-2\sqrt{30}}{5-6}$
$\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=\frac{11-2\sqrt{30}}{5-6}$
$\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=-11+2\sqrt{30}$

Nilai $a=-11$ dan $b=2$ maka $ab=-22$

$\therefore$ Pilihan yang sesuai ialah $(A)\ -22$

2. Soal SIMAK UI 2015 Kode 567 [Soal Lengkap Disini]Bentuk sederhana dari $\frac{\sqrt{143}+\sqrt{165}+\sqrt{195}+13}{\sqrt{11}+2\sqrt{13}+\sqrt{15}}$ adalah$\cdots$
$(A)\ \frac{1}{2} \left( \sqrt{15} + \sqrt{13} \right )$
$(B)\ \frac{1}{2} \frac{\left( \sqrt{15} - \sqrt{13} \right )}{\left( \sqrt{15} + \sqrt{13} \right )}$
$(C)\ \frac{1}{2} \left( \sqrt{15} - \sqrt{11} \right )$
$(D)\ \frac{1}{2} \left( \sqrt{15} + \sqrt{11} \right )$
$(E)\ \frac{\left( \sqrt{15} - \sqrt{11} \right )}{\left( \sqrt{15} + \sqrt{11} \right )}$Alternatif Pembahasan:

Hint

$\frac{\sqrt{143}+\sqrt{165}+\sqrt{195}+13}{\sqrt{11}+2\sqrt{13}+\sqrt{15}}$
$=\frac{\sqrt{11 \cdot 13}+\sqrt{11 \cdot 15}+\sqrt{13 \cdot 15}+\sqrt{13 \cdot 13}}{\sqrt{11}+\sqrt{13}+\sqrt{13}+\sqrt{15}}$
$=\frac{\left( \sqrt{13} + \sqrt{11} \right )\left( \sqrt{15} + \sqrt{13} \right )}{\sqrt{11} + \sqrt{13}+\sqrt{13} + \sqrt{15}}$
$=\frac{\left( \sqrt{13} + \sqrt{11} \right )\left( \sqrt{15} + \sqrt{13} \right )}{(\sqrt{11} + \sqrt{13})+(\sqrt{13} + \sqrt{15})} \times \frac{\left( \sqrt{13} - \sqrt{11} \right )\left( \sqrt{15} - \sqrt{13} \right )}{\left( \sqrt{13} - \sqrt{11} \right )\left( \sqrt{15} - \sqrt{13} \right )}$
$=\frac{\left( 13-11 \right )\left( 15 - 13 \right )}{2(\sqrt{15} - \sqrt{13})+2(\sqrt{13} - \sqrt{11})}$
$=\frac{4}{2\sqrt{15}-2\sqrt{11}}$
$=\frac{2}{\sqrt{15}-\sqrt{11}}$
$=\frac{2}{\sqrt{15}-\sqrt{11}} \times \frac{\sqrt{15}+\sqrt{11}}{\sqrt{15}+\sqrt{11}}$
$=\frac{2}{4} (\sqrt{15}+\sqrt{11})$
$=\frac{1}{2} (\sqrt{15}+\sqrt{11})$

$\therefore$ Pilihan yang sesuai ialah $(D)\ \frac{1}{2} \left( \sqrt{15} + \sqrt{11} \right )$

3. Soal UM UGM 2005 Kode 821Jika $\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}$ maka $\frac{1}{a}+\frac{1}{b}=\cdots$
$(A)\ 25$
$(B)\ 20$
$(C)\ 15$
$(D)\ 10$
$(E)\ 5$Alternatif Pembahasan:

Hint

$\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}$
$\sqrt{0,3+\sqrt{0,08}}$
$=\sqrt{\frac{3}{10}+\sqrt{\frac{8}{100}}}$
$=\sqrt{\frac{3}{10}+\sqrt{4 \cdot \frac{2}{100}}}$
$=\sqrt{\frac{3}{10}+2\sqrt{\frac{2}{100}}}$
$=\sqrt{\frac{1}{10}+\frac{2}{10}+2\sqrt{\frac{1}{10} \cdot \frac{2}{10}}}$
$=\sqrt{\frac{1}{10}}+\sqrt{\frac{2}{10}}$

Nilai $a=\frac{1}{10}$ dan $b=\frac{2}{10}$
$\frac{1}{a}+\frac{1}{b}=\frac{10}{1}+\frac{10}{2}=15$

$\therefore$ Pilihan yang sesuai ialah $(C)\ 15$

4.Soal SPMB 2007 Kode 341Jika dirasionalkan maka $1+\frac{1}{\sqrt{2}}+\frac{1}{1-\sqrt{2}}=\cdots$
$(A)\ -1-\frac{1}{2}\sqrt{2}$
$(B)\ -1-\sqrt{2}$
$(C)\ -\frac{1}{2}\sqrt{2}$
$(D)\ \frac{1}{2}\sqrt{2}$
$(E)\ 2+\frac{1}{2}\sqrt{2}$Alternatif Pembahasan:

Hint

Jika kita rasionalkan satu persatu, maka akan kita peroleh;
$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$
$\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1}{2}\sqrt{2}$

$\frac{1}{1-\sqrt{2}}=\frac{1}{1-\sqrt{2}} \cdot \frac{1+\sqrt{2}}{1+\sqrt{2}}$
$\frac{1}{1-\sqrt{2}}=\frac{1+\sqrt{2}}{1-2}$
$\frac{1}{1-\sqrt{2}}=\frac{1+\sqrt{2}}{-1}$
$\frac{1}{1-\sqrt{2}}=-1-\sqrt{2}$

Soal: $1+\frac{1}{\sqrt{2}}+\frac{1}{1-\sqrt{2}}$
$=1+\frac{1}{2}\sqrt{2}-1-\sqrt{2}$
$=-\frac{1}{2}\sqrt{2}$

$\therefore$ Pilihan yang sesuai ialah $(C)\ -\frac{1}{2}\sqrt{2}$

5. Soal SIMAK UI 2009 Kode 912 [Soal Lengkap Disini]Nilai dari $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots++\frac{1}{\sqrt{63}+\sqrt{64}}=\cdots$
$(A)\ 10$
$(B)\ 9$
$(C)\ 8$
$(D)\ 7$
$(E)\ 6$Alternatif Pembahasan:

Hint

Untuk menuntaskan soal ini dibutuhkan sedikit kreasi, yaitu dengan merasionalkan penyebut setiap suku;
$\frac{1}{1+\sqrt{2}}=-1+\sqrt{2}$
$\frac{1}{\sqrt{2}+\sqrt{3}}=-\sqrt{2}+\sqrt{3}$
$\frac{1}{\sqrt{3}+\sqrt{4}}=-\sqrt{3}+\sqrt{4}$
$\vdots$
$\frac{1}{\sqrt{62}+\sqrt{63}}=-\sqrt{62}+\sqrt{63}$
$\frac{1}{\sqrt{63}+\sqrt{64}}=-\sqrt{63}+\sqrt{64}$

$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots++\frac{1}{\sqrt{63}+\sqrt{64}}$
$=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-\cdots-\sqrt{62}+\sqrt{63}-\sqrt{63}+\sqrt{64}$
$=-1+\sqrt{64}$
$=-1+8=7$

$\therefore$ Pilihan yang sesuai ialah $(D)\ 7$

6. Soal UM UGM 2013 Kode 251$\frac{\sqrt{18}-\sqrt{12}}{\sqrt{18}+\sqrt{12}}+\frac{5}{1+\sqrt{6}}=\cdots$
$(A)\ \sqrt{6}$
$(B)\ 1-\sqrt{6}$
$(C)\ \sqrt{2}+\sqrt{3}$
$(D)\ 4-\sqrt{6}$
$(E)\ 5-2\sqrt{6}$Alternatif Pembahasan:

Hint

$\frac{\sqrt{18}-\sqrt{12}}{\sqrt{18}+\sqrt{12}}+\frac{5}{1+\sqrt{6}}$
$=\frac{\sqrt{9 \cdot 2}-\sqrt{4 \cdot 3}}{\sqrt{9 \cdot 2}+\sqrt{4 \cdot 3}}+\frac{5}{1+\sqrt{6}}$
$=\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}+\frac{5}{1+\sqrt{6}}$
$=\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} \cdot \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}+\frac{5}{1+\sqrt{6}} \cdot \frac{1-\sqrt{6}}{1-\sqrt{6}}$
$=\frac{18+12-12\sqrt{6}}{18-12}+\frac{5(1-\sqrt{6})}{1-6}$
$=\frac{30-12\sqrt{6}}{6}+\frac{5(1-\sqrt{6})}{-5}$
$=5-2\sqrt{6}-1+\sqrt{6}$
$=4-\sqrt{6}$

$\therefore$ Pilihan yang sesuai ialah $(D)\ 4-\sqrt{6}$

7. Soal UM UGM 2017 Kode 723Jika $r=\frac{20\sqrt{2}-25}{(10+20\sqrt{2})(2-\sqrt{2})}$, maka $(4r-2)^{2}=\cdots$
$(A)\ 5$
$(B)\ 4$
$(C)\ 3$
$(D)\ 2$
$(E)\ 1$Alternatif Pembahasan:

Hint

Salah satu trik menuntaskan duduk masalah matematika ialah kerjakan apa yang dapat dikerjakan hingga ketemu apa yang diharapkan. Seperti soal diatas diketahui $r$ dengan bentuk yang belum sederhana, mungkin dapat kita sederhanakan terlebih dahulu;
$r=\frac{20\sqrt{2}-25}{(10+20\sqrt{2})(2-\sqrt{2})}$
$r=\frac{20\sqrt{2}-25}{20-10\sqrt{2}+40\sqrt{2}-40}$
$r=\frac{20\sqrt{2}-25}{30\sqrt{2}-20}$
$r=\frac{5(4\sqrt{2}-5)}{10(3\sqrt{2}-2)}$
$r=\frac{1}{2} \cdot \frac{4\sqrt{2}-5}{3\sqrt{2}-2}$
$r=\frac{1}{2} \cdot \frac{4\sqrt{2}-5}{3\sqrt{2}-2} \cdot \frac{3\sqrt{2}+2}{3\sqrt{2}+2}$
$r=\frac{1}{2} \cdot \frac{24+8\sqrt{2}-15\sqrt{2}-10}{18-4}$
$r=\frac{1}{2} \cdot \frac{14-7\sqrt{2}}{14}$
$r=\frac{1}{2} \cdot \frac{7}{14} (2-\sqrt{2})$
$r=\frac{1}{4} (2-\sqrt{2})$

$(4r-2)^{2}=\left(4 \cdot \frac{1}{4} (2-\sqrt{2}) - 2 \right)^{2}$
$(4r-2)^{2}=\left(2-\sqrt{2} - 2 \right)^{2}$
$(4r-2)^{2}=\left(-\sqrt{2}\right)^{2}$
$(4r-2)^{2}=2$

$\therefore$ Pilihan yang sesuai ialah $(D)\ 2$

8. Soal SBMPTN 2015 Kode 634 [Soal Lengkap Disini]Jika $\sqrt[4]{a}+\sqrt[4]{9}=\frac{1}{2-\sqrt{3}}$, maka $a=\cdots$
$(A)\ 2-\sqrt{2}$
$(B)\ 2$
$(C)\ 2+\sqrt{2}$
$(D)\ 8$
$(E)\ 16$Alternatif Pembahasan:

Hint

$\sqrt[4]{a}+\sqrt[4]{9}=\frac{1}{2-\sqrt{3}}$
$\sqrt[4]{a}+9^{\frac{1}{4}}=\frac{1}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}$
$\sqrt[4]{a}+3^{\frac{1}{2}}=2+\sqrt{3}$
$\sqrt[4]{a}=2+\sqrt{3}-3^{\frac{1}{2}}$
$\sqrt[4]{a}=2$
$a=2^{4}=16$

$\therefore$ Pilihan yang sesuai ialah $(E)\ 16$

9. Soal UM UNDIP 2010 Kode 102 [Soal Lengkap Disini]Bentuk Sederhana dari $\sqrt{\frac{\sqrt{41}+4}{\sqrt{41}-4}}-\sqrt{\frac{\sqrt{41}-4}{\sqrt{41}+4}}$ adalah$\cdots$
$(A)\ -\frac{8}{5}$
$(B)\ 0$
$(C)\ \frac{16}{5}$
$(D)\ \frac{8}{5}$
$(E)\ 5\sqrt{41}$Alternatif Pembahasan:

Hint

$\sqrt{\frac{\sqrt{41}+4}{\sqrt{41}-4}}-\sqrt{\frac{\sqrt{41}-4}{\sqrt{41}+4}}$
$=\sqrt{\frac{\sqrt{41}+4}{\sqrt{41}-4} \cdot \frac{\sqrt{41}+4}{\sqrt{41}+4}}-\sqrt{\frac{\sqrt{41}-4}{\sqrt{41}+4} \cdot \frac{\sqrt{41}-4}{\sqrt{41}-4}}$
$=\sqrt{\frac{\left (\sqrt{41}+4\right )^{2}}{41-16}}-\sqrt{\frac{\left (\sqrt{41}-4\right )^{2}}{41-16}}$
$=\sqrt{\frac{\left (\sqrt{41}+4\right )^{2}}{25}}-\sqrt{\frac{\left (\sqrt{41}-4\right )^{2}}{25}}$
$=\frac{\sqrt{41}+4}{5}-\frac{\sqrt{41}-4}{5}$
$=\frac{\sqrt{41}+4-\sqrt{41}+4}{5}$
$=\frac{8}{5}$

$\therefore$ Pilihan yang sesuai ialah $(D)\ \frac{8}{5}$

10. Soal USM STIS 2017$\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\cdots$
$(A)\ 2\sqrt{3}$
$(B)\ \sqrt{10}$
$(C)\ 2\sqrt{2}$
$(D)\ \sqrt{11}$
$(E)\ 3\sqrt{2}$Alternatif Pembahasan:

Hint

Untuk menuntaskan soal ini, kita coba usahakan bentuk akar $\sqrt{3-\sqrt{5}}$ atau bentuk akar $\sqrt{3+\sqrt{5}}$ setidaknya seolah-olah dengan $\sqrt{(a+b)-2\sqrt{ab}}$ atau $\sqrt{(a+b)+2\sqrt{ab}}$.

$\sqrt{3-\sqrt{5}}$
$=\sqrt{3-2\sqrt{\frac{5}{4}}}$
$=\sqrt{\frac{5}{2}+\frac{1}{2}-2\sqrt{\frac{5}{4}}}$
$=\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}$

$\sqrt{3+\sqrt{5}}$
$=\sqrt{3+2\sqrt{\frac{5}{4}}}$
$=\sqrt{\frac{5}{2}+\frac{1}{2}+2\sqrt{\frac{5}{4}}}$
$=\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}$

$\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}$
$=\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}+\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}$
$=2\sqrt{\frac{5}{2}}$
$=2\sqrt{\frac{10}{4}}$
$=2 \cdot \frac{1}{2} \sqrt{10}$
$=\sqrt{10}$

$\therefore$ Pilihan yang sesuai ialah $(B)\ \sqrt{10}$

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