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# $\sqrt[n]{a}$ dibaca akar pangkat

bilangan

= indeks atau tingkat akar

= bilangan yang diambil akarnya atau bilangan pokok

# Apabila indeksnya 2, maka indeksnya dihilangkan sehingga penulisannya $\sqrt{a}$

# Menyederhanakan bentuk akar dapat dilakukan dengan cara:

menjadikan dalam bentuk perkalian dua buah akar pangkat bilangan

    $\sqrt{m}=\sqrt{a}\: \cdot \: \sqrt{b}$
dengan syarat salah satu akar pangkat bilangannya dapat di akarkan, contoh
$\sqrt{75}=\sqrt{{\color{Red} 25}}\: \cdot \: \sqrt{3}$
   $=\: {\color{Red} 5}\: \sqrt{3}$

# Penjumlahan dan pengurangan, jika akar-akarnya sejenis
    $a\: \sqrt[n]{c}\: \pm \: b\: \sqrt[n]{c}=\: \left ( a\: \pm\: b \right )\sqrt[n]{c}$
Contoh:

   $6\: \sqrt[3]{2}\: +\: 5\: \sqrt[3]{2}=\: 11\: \sqrt[3]{2}$

# Perkalian akar, jika akar-akarnya memiliki indeks yang sama
$x\: \sqrt[n]{a}\: \cdot \: y\: \sqrt[n]{b}=xy\: \sqrt[n]{ab}$

# $\sqrt{\left ( a+b \right )+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}$
Contoh:

   $\sqrt{8+2\sqrt{15}}=\sqrt{\left ( 3+5 \right )+2\sqrt{3\cdot 5}}$
   $=\sqrt{3}+\sqrt{5}$

# $\sqrt{\left ( a+b \right )-2\sqrt{ab}}=\sqrt{a}-\sqrt{b},\: \: a > b$
    Contoh:

$\sqrt{12-2\sqrt{35}}=\sqrt{\left ( 7+5 \right )-\sqrt{7\cdot 5}}$
$=\sqrt{7}-\sqrt{5}$

# Merasionalkan penyebut pecahan
   $\frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\: x\: \frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{b}=\frac{a}{b}\sqrt{b}$

   Contoh:
   $\frac{9}{2\sqrt{6}}=\frac{9}{2\sqrt{6}}\: x\: \frac{\sqrt{6}}{\sqrt{6}}$
              $=\frac{9\sqrt{6}}{2\cdot \sqrt{36}}$
$=\frac{9\sqrt{6}}{2\cdot 6}$
   $=\frac{3}{4}\sqrt{6}$
   $\frac{a}{c+\sqrt{b}}=\frac{a}{c+\sqrt{b}}\: x\: \frac{c-\sqrt{b}}{c-\sqrt{b}}$

     Contoh:
   $\frac{16}{3+\sqrt{5}}=\frac{16}{3+\sqrt{5}}\: x\: \frac{3-\sqrt{5}}{3-\sqrt{5}}$
   $=\frac{16\left ( 3-\sqrt{5} \right )}{9-3\sqrt{5}+3\sqrt{5}-\sqrt{25}}$
   $=\frac{16\left ( 3-\sqrt{5} \right )}{9-5}$
   $=\frac{16\left ( 3-\sqrt{5} \right )}{4}$
   $=4\left ( 3-\sqrt{5} \right )$

   $\frac{a}{c-\sqrt{b}}=\frac{a}{c-\sqrt{b}}\: x\: \frac{c+\sqrt{b}}{c+\sqrt{b}}$

   Contoh:
$\frac{2\sqrt{3}}{4-2\sqrt{3}}=\frac{2\sqrt{3}}{4-2\sqrt{3}}\: x\: \frac{4+2\sqrt{3}}{4+2\sqrt{3}}$
   $=\frac{2\sqrt{3}\left ( 4+2\sqrt{3} \right )}{\left ( 4-2\sqrt{3} \right )\left ( 4+2\sqrt{3} \right )}$
   $=\frac{8\sqrt{3}+4\sqrt{9}}{16+8\sqrt{3}-8\sqrt{3}-4\sqrt{9}}$
$=\frac{8\sqrt{3}+4\cdot 3}{16-4\cdot 3}$
$=\frac{8\sqrt{3}+12}{16-12}$
$=\frac{8\sqrt{3}+12}{4}$
$=2\sqrt{3}+3$

SOAL DAN PEMBAHASAN

1. UN 2007 P14
    Bentuk sederhana dari $\sqrt{8}\: +\: \sqrt{75}\; -\: \left ( \sqrt{32}\: +\: \sqrt{243} \right )\mathrm{adalah} .....$
    A. $2\sqrt{2}\: +\: 14\sqrt{3}$
B. $-2\sqrt{2}\: -\: 4\sqrt{3}$
C. $-2\sqrt{2}\: +\: 14\sqrt{3}$
D. $-2\sqrt{2}\: +\: 4\sqrt{3}$
E . $2\sqrt{2}\: -\: 4\sqrt{3}$

   Pembahasan:
$\sqrt{8}\: +\: \sqrt{75}\: -\: \left ( \sqrt{32}\: +\: \sqrt{243} \right )$
   $=\sqrt{4}\: \cdot \: \sqrt{2}\: +\: \sqrt{25}\: \cdot \: \sqrt{3}\: -\: \left ( \sqrt{16} \: \cdot \sqrt{2}+\sqrt{81}\: \cdot \: \sqrt{3}\right )$
$=2 \sqrt{2}\: +\: 5 \sqrt{3}\: -\: \left ( 4 \sqrt{2}+9 \sqrt{3}\right )$
$=2\sqrt{2}\: -\:4\sqrt{2}\: +\: 5\sqrt{3}\: -\: 9\sqrt{3}$
$= -2\sqrt{2}\: -\: 4\sqrt{3}$

Jawaban: ___________________________________ (B)

2. UN 2012 D74
Bentuk sederhana dari $\frac{\sqrt{2}+3\sqrt{5}}{\sqrt{2}-\sqrt{5}}$ adalah .....
    A. $\frac{1}{3}\left ( 17-4\sqrt{10} \right )$
B. $-\frac{2}{3}\left ( 15+4\sqrt{10} \right )$
C. $\frac{2}{3}\left ( 15-4\sqrt{10} \right )$
D. $\frac{1}{3}\left ( 17-4\sqrt{10} \right )$
E. $-\frac{1}{3}\left ( 17+4\sqrt{10} \right )$

Pembahasan:

Karena penyebutnya berbentuk akar maka kita rasionalkan, dikalikan dengan bentuk sekawannya

   $\frac{\sqrt{2}+3\sqrt{5}}{\sqrt{2}-\sqrt{5}}$

$=\frac{\sqrt{2}+3\sqrt{5}}{\sqrt{2}-\sqrt{5}}\: x\: \frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}$
$=\frac{\left (\sqrt{2}+3\sqrt{5} \right )\left ( \sqrt{2}+\sqrt{5} \right )}{\left ( \sqrt{2}-\sqrt{5} \right )\left ( \sqrt{2}+\sqrt{5} \right )}$
$=\frac{\sqrt{4}+\sqrt{10}+3\sqrt{10}+3\sqrt{25}}{\sqrt{4}+\sqrt{10}-\sqrt{10}-\sqrt{25}}$
$=\frac{2+4\sqrt{10}+3\: \cdot \: 5}{2-5}$
$=\frac{17+4\sqrt{10}}{-3}$
$=-\frac{1}{3}\left ( 17+4\sqrt{10} \right )$

Jawaban: ___________________________________ (E)

3. USM STIS 2014
     Bentuk sederhana dari $\frac{\sqrt{15}}{\sqrt{10}-\sqrt{6}}$ adalah .....
    A. $\frac{5}{4}\sqrt{10}+\frac{3}{4}\sqrt{6}$
B. $\frac{5}{4}\sqrt{6}-\frac{3}{4}\sqrt{10}$
    C. $\frac{5}{4}\sqrt{6}+\frac{3}{4}\sqrt{10}$
D. $\frac{3}{4}\sqrt{10}-\frac{5}{4}\sqrt{10}$
E.   $\frac{3}{4}\sqrt{10}-\frac{5}{4}\sqrt{6}$

Pembahasan:

Karena penyebutnya berbentuk akar maka dirasionalkan

   $\frac{\sqrt{15}}{\sqrt{10}-\sqrt{6}}$
$=\frac{\sqrt{15}}{\sqrt{10}-\sqrt{6}}\: x\: \frac{\sqrt{10}+\sqrt{6}}{\sqrt{10}+\sqrt{6}}$
$=\frac{\sqrt{15}\left ( \sqrt{10}+\sqrt{6} \right )}{\left (\sqrt{10}-\sqrt{6} \right )\left ( \sqrt{10} +\sqrt{6}\right )}$
$=\frac{\sqrt{150}+\sqrt{90}}{\sqrt{100}+\sqrt{60}-\sqrt{60}-\sqrt{36}}$
$=\frac{\sqrt{25}\: \cdot \: \sqrt{6}+\sqrt{9}\: \cdot \: \sqrt{10}}{10-6}$
$=\frac{5\sqrt{6}+3 \sqrt{10}}{4}$
$=\frac{5}{4}\sqrt{6}+\frac{3}{4}\sqrt{10}$

Jawaban: ___________________________________ (C)

4. SIMAK UI 2009 Kode 912

    $\sqrt{3+2\sqrt{2}}-\sqrt{2}\: =\: .....$
   A. $4\sqrt{2}$
   B . $3+\sqrt{2}$
   C. $\sqrt{2}$
   D.

Pembahasan:

$\sqrt{3+2\sqrt{2}}-\sqrt{2}$
$=\sqrt{\left ( 2+1 \right )+2\sqrt{2\: \cdot \: 1}}-\sqrt{2}$
$=\sqrt{2}+\sqrt{1}-\sqrt{2}$

Jawaban: ___________________________________ (D)

5. SPMB 2006 Kode 510
Jika bilangan bulat

dan

memenuhi $\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a\: +\: b\sqrt{30}$ , maka $ab=\: .....$
A.

Pembahasan:

$\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a\: +\: b\sqrt{30}$
    $\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}\: x\: \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}=a\: +\: b\sqrt{30}$
   $\frac{\left ( \sqrt{5}-\sqrt{6} \right )\left ( \sqrt{5}-\sqrt{6} \right )}{\left ( \sqrt{5}+\sqrt{6} \right )\left ( \sqrt{5}-\sqrt{6} \right )}=a\: +\: b\sqrt{30}$
$\frac{\sqrt{25}-\sqrt{30}-\sqrt{30}+\sqrt{36}}{\sqrt{25}-\sqrt{30}+\sqrt{30}-\sqrt{36}}=a\: +\: b\sqrt{30}$
   $\frac{5-2\sqrt{30}+6}{5-6}=a\: +\: b\sqrt{30}$
   $\frac{11-2\sqrt{30}}{-1}=a\: +\: b\sqrt{30}$
$-11+2\sqrt{30}=a\: +\: b\sqrt{30}$ , maka diperoleh

maka $ab=\left ( -11 \right )\left ( 2 \right )$

Jawaban: ___________________________________ (A)

6. UM UGM 2005 Kode 821

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

A.

Pembahasan:

   $\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}$
   $\sqrt{0,3+\sqrt{4}\: \cdot \: \sqrt{0,02}}=\sqrt{a}+\sqrt{b}$
   $\sqrt{0,3+2 \sqrt{0,02}}=\sqrt{a}+\sqrt{b}$
$\sqrt{\left ( 0,2+0,1 \right )+2 \sqrt{0,2\: \cdot \: 0,1}}=\sqrt{a}+\sqrt{b}$
   $\sqrt{0,2}+\sqrt{0,1}=\sqrt{a}+\sqrt{b}$ , maka diperoleh
$a=0,2=\frac{2}{10}$
$b=0,1=\frac{1}{10}$
maka $\frac{1}{a}+\frac{1}{b}=\frac{1}{\frac{2}{10}}+\frac{1}{\frac{1}{10}}$
   $=\frac{10}{2}+\frac{10}{1}$

Jawaban: ___________________________________ (C)

7. 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.

Pembahasan:

Karena penyebutnya bentuk akar maka dirasionalkan

$\frac{1}{1+\sqrt{2}}=\frac{1}{1+\sqrt{2}}\: x\: \frac{1-\sqrt{2}}{1-\sqrt{2}}$
   $= \frac{1-\sqrt{2}}{\left ( 1+\sqrt{2} \right )\left ( 1-\sqrt{2} \right )}$
   $= \frac{1-\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{4}}$
   $= \frac{1-\sqrt{2}}{1-2}$
$= \frac{1-\sqrt{2}}{-1}$
   $= \sqrt{2}-1$
$\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{1}{\sqrt{2}+\sqrt{3}}\: x\: \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$
$= \frac{\sqrt{2}-\sqrt{3}}{\left ( \sqrt{2}+\sqrt{3} \right )\left ( \sqrt{2}-\sqrt{3} \right )}$
   $= \frac{\sqrt{2}-\sqrt{3}}{\sqrt{4}-\sqrt{6}+\sqrt{6}-\sqrt{9}}$
   $= \frac{\sqrt{2}-\sqrt{3}}{2-3}$
$= \frac{\sqrt{2}-\sqrt{3}}{-1}$
   $= \sqrt{3}-\sqrt{2}$
$\frac{1}{\sqrt{3}+\sqrt{4}}=\frac{1}{\sqrt{3}+\sqrt{4}}\: x\: \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}$
$= \frac{\sqrt{3}-\sqrt{4}}{\left ( \sqrt{3}+\sqrt{4} \right )\left ( \sqrt{3}-\sqrt{4} \right )}$
   $= \frac{\sqrt{3}-\sqrt{4}}{\sqrt{9}-\sqrt{12}+\sqrt{12}-\sqrt{16}}$
   $= \frac{\sqrt{3}-\sqrt{4}}{3-4}$
$= \frac{\sqrt{3}-\sqrt{4}}{-1}$
   $= \sqrt{4}-\sqrt{3}$

$\frac{1}{\sqrt{63}+\sqrt{64}}=\frac{1}{\sqrt{63}+\sqrt{64}}\: x\: \frac{\sqrt{63}-\sqrt{64}}{\sqrt{63}-\sqrt{64}}$
   $= \frac{\sqrt{63}-\sqrt{64}}{\left ( \sqrt{63}+\sqrt{64} \right )\left ( \sqrt{63}-\sqrt{64} \right )}$
$= \frac{\sqrt{63}-\sqrt{64}}{\sqrt{3969}-\sqrt{40932}+\sqrt{4032}-\sqrt{4096}}$
   $= \frac{\sqrt{63}-\sqrt{64}}{63-64}$
   $= \frac{\sqrt{63}-\sqrt{64}}{-1}$
   $= \sqrt{64}-\sqrt{63}$

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

Jawaban: ___________________________________ (D)

8. SBMPTN 2015 Kode 621
Jika $\sqrt[3]{a}+\sqrt[4]{9}=\frac{1}{2-\sqrt{3}}$ , maka

A. $2-\sqrt{3}$
B.

C. $2+\sqrt{3}$
D.

Pembahasan:

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

$\sqrt[3]{a}+\sqrt[4]{3^{2}}=\frac{1}{2-\sqrt{3}}\: x\: \frac{2+\sqrt{3}}{2+\sqrt{3}}$
   $\sqrt[3]{a}+\sqrt{3}= \frac{2+\sqrt{3}}{\left (2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right )}$
   $\sqrt[3]{a}+\sqrt{3}= \frac{2+\sqrt{3}}{4+2\sqrt{3}-2\sqrt{3}-\sqrt{9}}$
$\sqrt[3]{a}+\sqrt{3}= \frac{2+\sqrt{3}}{4-3}$
$\sqrt[3]{a}+\sqrt{3}= 2+\sqrt{3}$ , maka diperoleh

$\sqrt[3]{a}= 2$      kedua ruas dipangkatkan 3
$\left ( a^{\frac{1}{3}} \right )^{3}=2^{3}$

Jawaban: ___________________________________ (D)

M@th

Rabu, 09 Agustus 2017

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