Soal dan Pembahasan Bentuk Akar

Soal 1: UM STIS 2017

$\sqrt{3- \sqrt{5}} + \sqrt{3+ \sqrt{5}} = . . .$ 

(A) $2 \sqrt{3}$
(B) $\sqrt{10}$
(C) $2 \sqrt{2}$
(D) $\sqrt{11}$ 

(E) $3 \sqrt{2}$

PEMBAHASAN:

Pertama kita akan ubah bentuk akar $\sqrt{a \pm \sqrt{b}}$ menjadi $\sqrt{(a+b) \pm 2\sqrt{ab}} = \sqrt{a} \pm \sqrt{b}$,maka dari soal

$\begin{aligned} \sqrt{3 \pm \sqrt{5}} &= \sqrt{3 \pm 2 \sqrt{\frac{5}{4}}} \\ &= \sqrt{3 \pm 2 \sqrt{\frac{5}{4}}}\\ &= \sqrt{\frac{5}{2} + \frac{1}{2} \pm 2 \sqrt{\frac{5}{2}.\frac{1}{2}}}\\ \sqrt{3 \pm \sqrt{5}} &= \sqrt{\frac{5}{2}} \pm \sqrt{\frac{1}{2}},\ atau \\ \sqrt{3 +\sqrt{5}} &= \sqrt{\frac{5}{2}} + \sqrt{\frac{1}{2}},\ dan \\ \sqrt{3 -\sqrt{5}} &= \sqrt{\frac{5}{2}} - \sqrt{\frac{1}{2}} \\ \end{aligned}$

maka

$\begin{aligned} \sqrt{3 -\sqrt{5}} + \sqrt{3 +\sqrt{5}} &= \sqrt{\frac{5}{2}} - \sqrt{\frac{1}{2}} + \sqrt{\frac{5}{2}} + \sqrt{\frac{1}{2}} \\ &= 2 \sqrt{\frac{5}{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ &= 2\sqrt{\frac{10}{4}}= 2. \frac{1}{2} \sqrt{10}\\ \sqrt{3 -\sqrt{5}} + \sqrt{3 +\sqrt{5}} &= \sqrt{10}\end{aligned}$

Jadi, pilihan jawabannya adalah (B), $\sqrt{10}$

Soal 2: UM UNDIP 2010 Kode 102

Bentuk sederhana dari $\sqrt{\frac{\sqrt{41}+4}{\sqrt{41}-4}} -\sqrt{\frac{\sqrt{41}-4}{\sqrt{41}+4}}$ adalah . . .

(A) $- \frac{8}{5}$
(B) $0$
(C) $\frac{16}{5}$
(D) $\frac{8}{5}$ 

(E) $5 \sqrt{41}$

PEMBAHASAN:

$\sqrt{\frac{\sqrt{41}+4}{\sqrt{41}-4}} -\sqrt{\frac{\sqrt{41}-4}{\sqrt{41}+4}}$

= $\sqrt{\frac{\sqrt{41}+4}{\sqrt{41}-4} \times \frac{\sqrt{41}+4}{\sqrt{41}+4}} -\sqrt{\frac{\sqrt{41}-4}{\sqrt{41}+4} \times \frac{\sqrt{41}-4}{\sqrt{41}-4}}$ 

= $\sqrt{\frac{(\sqrt{41}+4)^2}{41-16}} -\sqrt{\frac{(\sqrt{41}-4)^2}{41-16}}$ 

= $\sqrt{\frac{(\sqrt{41}+4)^2}{25}} -\sqrt{\frac{(\sqrt{41}-4)^2}{25}}$ 

= $\frac{\sqrt{41}+4}{5} - \frac{\sqrt{41}-4}{5}$ 

= $\frac{\sqrt{41}+4-\sqrt{41}+4}{5} = \frac{8}{5}$ 

Jadi, pilihan jawabannya adalah (D), $\frac{8}{5}$</

Soal 3: SBMPTN 2015 Kode 634/610

Jika $\sqrt[4]{a} +\sqrt[4]{9} = \frac{1}{2 - \sqrt{3}}$ maka $a = . . .$.

(A) $2 - \sqrt{2}$
(B) $2$
(C) $2 + \sqrt{2}$
(D) $8$ 

(E) $16$

PEMBAHASAN: 

$\begin{aligned} \sqrt[4]{a} +\sqrt[4]{9} &= \frac{1}{2 - \sqrt{3}} \\ \sqrt[4]{a} + 3^{2 \times \frac{1}{4}} &= \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} \\ \sqrt[4]{a} + \sqrt{3} &=2 + \sqrt{3}\\ \sqrt[4]{a} &= 2 \\ a &= 2^4 = 16\end{aligned}$

Jadi, pilihan jawabannya adalah (E), $16$</

Soal 4: UM UGM 2017 Kode 723

Jika $r = \frac{20 \sqrt{2} -25}{(10 + 20 \sqrt{2})(2 - \sqrt{2})}$ maka $(4r - 2)^2 = . . .$.

(A) $5$
(B) $4$
(C) $3$
(D) $2$ 

(E) $1$

PEMBAHASAN:

Pertama kita mengubah bentuk r ke dalam bentuk yang paling sederhana $\begin{aligned} r &= \frac{20 \sqrt{2} -25}{(10 + 20 \sqrt{2})(2 - \sqrt{2})} \\ r& = \frac{20 \sqrt{2} -25}{20 - 10 \sqrt{2}+40 \sqrt{2} -40}\\ & = \frac{20 \sqrt{2} -25}{30\sqrt{2} - 20}\\ & = \frac{5}{10}\left (\frac{4 \sqrt{2} -5}{3\sqrt{2} - 2} \right )\\ & = \frac{1}{2}\left (\frac{4 \sqrt{2} -5}{3\sqrt{2} - 2} \right ) \times \frac{3\sqrt{2}+2}{3\sqrt{2}+2}\\ & = \frac{1}{2}\ .\ \frac{24+8 \sqrt{2} - 15 \sqrt{2} -10}{18-4}\\ & = \frac{1}{2}\ .\ \frac{14-7 \sqrt{2} }{14}\\ & = \frac{1}{2}\ .\ \frac{7}{14} \ . \ (2- \sqrt{2})\\ r & = \frac{1}{4}(2- \sqrt{2})\ \end{aligned}$ 

maka nilai dari 

$\begin{aligned} (4r-2)^2 &= \left [4 \times \frac{1}{4}(2- \sqrt{2}) -2\right ]^2\\ & = (2 - \sqrt{2}-2)^2\\ & = (- \sqrt{2})^2 = 2\end{aligned}$

Jadi, pilihan jawabannya adalah (D), $2$</

Soal 5: UM UGM 2013 Kode 251

$\frac{\sqrt{18} - \sqrt{12}}{\sqrt{18} + \sqrt{12}} + \frac{5}{1+ \sqrt{6}} = . . .$ 

(A) $\sqrt{6}$
(B) $1 - \sqrt{6}$
(C) $\sqrt{2} + \sqrt{3}$
(D) $4 - \sqrt{6}$ 

(E) $5 - 2 \sqrt{6}$

PEMBAHASAN:

$\frac{\sqrt{18} - \sqrt{12}}{\sqrt{18} + \sqrt{12}} + \frac{5}{1+ \sqrt{6}}$

= $\frac{\sqrt{9.2} - \sqrt{4.3}}{\sqrt{9.2} + \sqrt{4.3}} + \frac{5}{1+ \sqrt{6}}$

= $\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \frac{5}{1+ \sqrt{6}}$

= $\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} \times \frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} + \frac{5}{1+ \sqrt{6}} \times \frac{1- \sqrt{6}}{1- \sqrt{6}}$

= $\frac{18+12-12 \sqrt{6}}{18-12} + \frac{5{(1- \sqrt{6})}}{1-6}$

= $\frac{30-12 \sqrt{6}}{6} + \frac{5(1- \sqrt{6})}{-5}$ 

= $5 - 2\sqrt{6} - 1 + \sqrt{6}$

= $4 - \sqrt{6}$

Jadi, pilihan jawabannya adalah (D), $4 - \sqrt{6}$</

 

Soal 6: SIMAK UI 2009 Kode 912

Nilai dari $\frac{1}{1+ \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + . . . + \frac{1}{\sqrt{63} + \sqrt{64}} = . . .$

(A) $10$
(B) $9$
(C) $8$
(D) $7$ 

(E) $6$

PEMBAHASAN:

Kita terlebih dahulu merasionalkan penyebut dari setiap suku.

$\Rightarrow \frac{1}{1+ \sqrt{2}} \times \frac{1- \sqrt{2}}{1- \sqrt{2}} = -1 + \sqrt{2}$

$\Rightarrow \frac{1}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2}- \sqrt{3}}{\sqrt{2} - \sqrt{3}} = -\sqrt{2} + \sqrt{3}$

$\Rightarrow \frac{1}{\sqrt{3} + \sqrt{4}} \times \frac{\sqrt{3}- \sqrt{4}}{\sqrt{3} - \sqrt{4}} = -\sqrt{3} + \sqrt{4}$

seterusnya

$\Rightarrow \frac{1}{\sqrt{62} + \sqrt{63}} \times \frac{\sqrt{62}- \sqrt{63}}{\sqrt{62} - \sqrt{63}} = -\sqrt{62} + \sqrt{63}$

$\Rightarrow \frac{1}{\sqrt{63} + \sqrt{64}} \times \frac{\sqrt{63}- \sqrt{64}}{\sqrt{63} - \sqrt{64}} = -\sqrt{63} + \sqrt{64}$

maka 

$\frac{1}{1+ \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + . . . + \frac{1}{\sqrt{63} + \sqrt{64}}$

= $-1 + \sqrt{2} - \sqrt{2} + \sqrt{3} + ... + \sqrt{63} - \sqrt{63} + \sqrt{64}$

= $-1 + \sqrt{64}$

= $-1 + 8 = 7$

Jadi, pilihan jawabannya adalah (D), $7$</

 

Soal 7: SPMB 2007 Kode 341

Jika dirasionalkan maka  $1 + \frac{1}{\sqrt{2}} + \frac{1}{1- \sqrt{2}} = . . .$

(A) $-1 - \frac{1}{2} \sqrt{2}$
(B) $-1 - \sqrt{2}$
(C) $- \frac{1}{2} \sqrt{2}$
(D) $\frac{1}{2}$ 

(E) $2 + \frac{1}{2} \sqrt{2}$

PEMBAHASAN:

Kita rasionalkan satu per satu kemudia kita jumlahkan:

$\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{1}{2} \sqrt{2}$

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

Jadi  

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

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

= $-\frac{1}{\sqrt{2}}\sqrt{2}$

Jadi, pilihan jawabannya adalah (C), $- \frac{1}{2} \sqrt{2}$</

Soal 8: SPMB 2007 Kode 341

Jika $\sqrt{0,3 + \sqrt{0,08}} = \sqrt{a} + \sqrt{b}$, maka $\frac{1}{a} + \frac{1}{b} = . . .$

(A) $25$
(B) $20$
(C) $15$
(D) $10$ 

(E) $5$

PEMBAHASAN:

Kita gunakan terorema $\sqrt{(a+b)+2\sqrt{ab}}=\sqrt{a} + \sqrt{b}$, maka

$\sqrt{0,3 + \sqrt{0,08}}$

= $\sqrt{\frac{3}{10} + \sqrt{\frac{8}{100}}}$

= $\sqrt{\frac{3}{10} + \sqrt{4.\frac{2}{100}}}$

= $\sqrt{\frac{3}{10} + 2\sqrt{\frac{2}{100}}}$

= $\sqrt{\frac{1}{10} + \frac{2}{10} + 2\sqrt{\frac{1}{10} \times \frac{2}{10}}}$

= $\sqrt{\frac{1}{10}} + \sqrt{\frac{2}{10}} $

maka nilai $a = \frac{1}{10}$ dan $b = \frac{2}{10}$, jadi

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

Jadi, pilihan jawabannya adalah (C), $15$</

Soal 9: SIMAK UI 2015 Kode 567

Bentuk sederhana dari $\frac{\sqrt{143} + \sqrt{165} + \sqrt{195} + 13}{\sqrt{11} + 2\sqrt{13} + \sqrt{15}}$ adalah . . .

(A) $\frac{1}{2}(\sqrt{15} + \sqrt{13})$
(B) $\frac{1}{2} \frac{\left (\sqrt{15}- \sqrt{13}  \right )}{\left (\sqrt{15} + \sqrt{13}  \right )}$
(C) $\frac{1}{2}(\sqrt{15}-\sqrt{11})$
(D) $\frac{1}{2}(\sqrt{15}+\sqrt{11})$ 

(E) $\frac{\left (\sqrt{15}- \sqrt{11}  \right )}{\left (\sqrt{15} + \sqrt{11}  \right )}$

Soal 10: SPMB 2006 Kode 510

Jika bilangan bulat a dan b memenuhi $\frac{\sqrt{5} - \sqrt{6}}{\sqrt{5} + \sqrt{6}} = a + b \sqrt{30}$ maka $ab = . . . $  

(A) $-22$
(B) $-22$
(C) $-9$
(D) $2$ 

(E) $13$

PENYELESAIAN:

$\begin{aligned} \frac{\sqrt{5} - \sqrt{6}}{\sqrt{5} + \sqrt{6}} &= \frac{\sqrt{5} - \sqrt{6}}{\sqrt{5} + \sqrt{6}} \times \frac{\sqrt{5} - \sqrt{6}}{\sqrt{5} - \sqrt{6}} \\ &= \frac{5+6-2 \sqrt{30}}{5-6}\\ &= -11 + 2\sqrt{30} \end{aligned}$

maka a = -11 dan b = 2, maka ab = -22

Jadi, pilihan jawabannya adalah (B), $-22$</