SIFAT SIFAT BENTUK AKAR 



Bentuk umum akar 

\(\sqrt[n]{a^{m}} = a^{\frac{m}{n}} \) 
\( \sqrt[n]{a} = a^{\frac{1}{n}}\) 
\( \sqrt{a^{m}} = a^{\frac{m}{2}} \) 
\( \sqrt {a} = a^{\frac{1}{2}}\) 

Penjumlahan dan pengurangan 

\( a\sqrt{c} + b\sqrt{c} = (a + b)\sqrt{c} \) 
\( a\sqrt{c} – b\sqrt{c} = (a - b)\sqrt{c} \) 

Perkalian dan pembagian 

\( \sqrt{a} \cdot \sqrt{a} = a \) 
\( \sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{ab}\) 
\( \sqrt[n]{a^{m}} \cdot \sqrt[n]{a^{p}} = \sqrt[n]{a^{m+p}} \) 
\( \sqrt[n]{\sqrt[p]{a}} = \sqrt[np]{a}\) 
\( \frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}} \) 

MERASIONALKAN PENYEBUT 

\(\frac {a} {\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac {\sqrt {b}}{\sqrt {b}} = \frac {a\sqrt {b}}{b} \)
\( \frac{\sqrt {a}} {\sqrt{b}} = \frac {\sqrt {a}} {\sqrt{b}} \times \frac {\sqrt{b}}{\sqrt{b}} = \frac{\sqrt {ab}}{b} \) 
\( \frac {c} {\sqrt{a} + \sqrt {b}} = \frac {c} {\sqrt{a} + \sqrt {b}} \times \frac {\sqrt{a} - \sqrt{b}} {\sqrt{a} - \sqrt{b}} = \frac {c(\sqrt{a} - \sqrt{b})} {a - b} \) 
\( \frac {\sqrt {a} + \sqrt {b}} {\sqrt {a} - \sqrt {b}} = \frac {\sqrt {a} + \sqrt {b}} {\sqrt {a} - \sqrt {b}} \times \frac {\sqrt {a} + \sqrt {b}} {\sqrt {a} + \sqrt {b}} = \frac {( \sqrt {a} + \sqrt {b} )^{2}} {a - b} \) 

PERSAMAAN BENTUK AKAR 

\( \sqrt{ (a + b ) + 2\sqrt{ab}} = \sqrt {a} + \sqrt {b}\), syarat : a > b > 0 
Bukti : 
\(\begin {eqnarray} \sqrt {(a + b) + 2\sqrt{ab}} &=& \sqrt { a + 2\sqrt{ab} + b} \\ &=& \sqrt {a + \sqrt {ab} + \sqrt {ab} + b } \\ &=& \sqrt{ \sqrt {a^{2}} + \sqrt{ab} + sqrt {ab} + \sqrt{b^{2}}} \\ &=& \sqrt{ \sqrt {a}(\sqrt{a} + \sqrt{b}) + \sqrt {b} (\sqrt{a} + \sqrt {b}) } \\ &=& \sqrt {(\sqrt {a} + \sqrt {b})(\sqrt {a} + \sqrt {b})} \\ &=& \sqrt {(\sqrt {\sqrt {a} + \sqrt {b}})} \\ &=& \sqrt {a} +\sqrt {b} \\ \end {eqnarray}\) 
\( \sqrt {(a + b) – 2\sqrt {ab}} = \sqrt {a} - \sqrt {b}\), syarat a > b > 0 
Bukti : 
\(\begin {eqnarray} \sqrt {(a + b) – 2\sqrt {ab}} &=& \sqrt {a - 2\sqrt{ab} + b} \\ &=& \sqrt {a - \sqrt{ab} - \sqrt {ab} + b} \\ &=& \sqrt {(\sqrt {a})^{2} -\sqrt{ab} -\sqrt {ab} + \sqrt{(b)^{2}}} \\ &=& \sqrt {\sqrt{a}(\sqrt{a} - \sqrt {b}) - \sqrt {b}(\sqrt {a} - \sqrt {b})} \\ &=& \sqrt {(\sqrt {a} - \sqrt {b})(\sqrt {a} - \sqrt {b})} \\ &=& \sqrt {(\sqrt {a} - \sqrt {b})^{2}} \\ &=& \sqrt {a} -\sqrt {b} \end {eqnarray}\)

CONTOH SOAL 

1. UM UNDIP TAHUN 2009 KODE 191 
Bentuk sederhana dari \( \frac{ a\sqrt{a} + b\sqrt{b} }{ \sqrt{a} + \sqrt {b} } = …  \) 
a. \( a + b - \sqrt {ab} \) 
 b. \( a – b + \sqrt {ab} \) 
c. \( a + b + \sqrt {ab} \) 
d. \( a – b - \sqrt {ab} \) 
e. \( -a – b - \sqrt {ab} \) 
PEMBAHASAN : 
\(\begin {eqnarray} \frac{ a\sqrt{a} + b\sqrt{b} }{\sqrt{a} + \sqrt {b} } \times \frac {\sqrt {a} - \sqrt{b} }{ \sqrt {a} - \sqrt{b}} &=& \frac {( a\sqrt {a} + b\sqrt{b} )( \sqrt{a} - \sqrt{b} )}{a - b} \\ &=& \frac { a^{2} – a\sqrt {ab} + b\sqrt {ab} – b^{2} }{a - b} \\ &=&\frac { a^{2} – b^{2} – (a – b (\sqrt {ab})) }{ a- b } \\ &=& \frac { (a + b)( a + b) - ( a – b ( \sqrt {ab} ) ) }{ a - b } \\ &=& \frac { (a - b)( a + b ) – (\sqrt {ab})) }{ a-b } \\ &=& a + b - \sqrt{ab} \end {eqnarray}\)
JAWABAN : A

2. UNM UNDIP TAHUN 2008 KODE 581 
Nilai dari \( \sqrt{9 + \sqrt {17}} - \sqrt{9 - \sqrt {17}} = … \) 
a. 1 
b. \(\sqrt{2}\) 
c. \( \sqrt {3} \) 
d. 2 
e. \( \sqrt{5} \) 
 PEMBAHASAN : 
 \(\begin {eqnarray} \sqrt {9 + \sqrt {17}} - \sqrt {9 - \sqrt {17}} &=& x \\ ( \sqrt {9 + \sqrt {17}} - \sqrt {9 - \sqrt {17}} )^{2} = x^{2} \\ 9 + \sqrt {17} – 2\sqrt {( 9 + \sqrt{17} )( 9 - \sqrt{17} )} + 9 - \sqrt {17} &=& x^{2} \\ 18 – 2\sqrt {81 - 17} &=& x^{2} \\ 18 – 2\sqrt {64} &=& x^{2} \\ x^{2} &=& 18 – 16 \\ x^{2} &=& 4 \\ x &=& 2 \end {eqnarray}\) 
JAWABAN : D

3. UM UGM TAHUN 2013 KODE 251 
Bentuk sederhana \( \frac {\sqrt {18} - \sqrt {12} }{ \sqrt {18} + \sqrt {12} } + \frac { 5 }{ 1 + \sqrt {6} }=…\) 
 a. \( \sqrt {6} \) 
 b. \( 1 -\sqrt {6} \) 
c. \( \sqrt {2}+ \sqrt {3} \) 
d. \( 4 - \sqrt {6} \) 
e. \( 5 - 2\sqrt {6} \)
PEMBAHASAN : 
\ \(\begin {eqnarray} \frac {\sqrt {18} - \sqrt {12} }{ \sqrt {18} + \sqrt {12} } + \frac { 5 }{ 1 + \sqrt {6}} &=& \frac {\sqrt {18} - \sqrt {12} }{ \sqrt {18} + \sqrt {12} } \times \frac { \sqrt {18} - \sqrt {12} }{ \sqrt {18} - \sqrt {12} } + \frac { 5 }{ 1 + \sqrt {6}} \times \frac { 1 - \sqrt {6} } { 1 - \sqrt {6} } \\ &=& \frac { 18 – 2\sqrt{18 - 12} + 12 }{ 18 - 12 } + \frac { 5 – 5\sqrt {6} }{ 1 - 6 } \\ &=& \frac { 30 – 12\sqrt {6} }{ 6 } + ( - \frac { 5 - 5\sqrt{6} }{ 5 } ) \\ &=& \frac { 6(5 - 2\sqrt{6}) }{ 6 } + ( - \frac { 5(1 - \sqrt {6}) }{ 5 } ) \\ &=& 5 - 2\sqrt{6} - 1 - \sqrt{6} \\ &=& 4 - \sqrt {6} \end {eqnarray}\)
JAWABAN : D 

4. UNPAD 2010 KODE 041 
Bentuk \( \sqrt [4]{49 – 20\sqrt {6}} \) dapat disederhanakan menjadi … 
a. \( \sqrt {3 - }\sqrt {2} \) 
b. \( 5 -2\sqrt {6} \) 
c. \( 7 + 2\sqrt {6} \) 
d. \( \sqrt {6} - 1 \) 
e. \( \sqrt {5} - 1 \) 
PEMBAHASAN : 
 \(\begin {eqnarray} \sqrt [4]{49 – 20\sqrt {6}} &=& \sqrt {\sqrt {49 – 20\sqrt {6}}} \\ &=& \sqrt { \sqrt{ 49 – 2\sqrt {600} } } \\ &=& \sqrt { \sqrt { 25 +24 -2\sqrt {25 \cdot 24} } } \\ &=& \sqrt { 5 -\sqrt {24} } \\ &=& \sqrt {5 – 2\sqrt {6}} \\ &=& \sqrt {3 + 2 -\sqrt {3 \cdot 2}} \\ &=& \sqrt {3} - \sqrt {2} \end {eqnarray}\) 
JAWABAN : A 

5. SPMB 2002 
Jika \( \frac {\sqrt{2} - \sqrt{3}} {\sqrt{2} + \sqrt{3}} = a + b\sqrt{6} \), a dan b adalah bilangan bulat, maka \( a + b \)… 
a. -5 
b. -3 
c. -2 
d. 2 
e. 3 
PEMBAHASAN : 
\(\begin {eqnarray} \frac {\sqrt{2} - \sqrt{3}} {\sqrt{2} + \sqrt{3}} &=& a + b\sqrt{6} \\ \frac { \sqrt{2} - \sqrt{3} } { \sqrt{2} + \sqrt{3} } \times \frac { \sqrt{2} - \sqrt{3} } { \sqrt{2} - \sqrt{3} } &=& a + b\sqrt{6} \\ \frac { 2 - 2\sqrt{6} + 3 } { 2 - 3 } &=& a + b\sqrt{6} \\ \frac { 5 – 2\sqrt{6} } { -1 } &=& a + b\sqrt{6} \\ -5 + 2\sqrt{6} &=& a + b\sqrt{6} \\ \end {eqnarray}\) 
  • \(a = -5\)  
  • \(b = 2\)
  • \(a + b = -3\) 
JAWABAN : C 

6. UM UGM TAHUN 2005 DAN 2006 
Apabila \( \frac{\sqrt{8}}{\sqrt{5} - \sqrt{3}} \) dirasionalkan penyebutnya, maka … 
a. \( \sqrt{10} + \sqrt{6} \)
b. \( \sqrt{10} + \sqrt{3} \) 
c. \( \sqrt{10} - \sqrt{6} \) 
d. \( 2\sqrt{5} - \sqrt{3} \) 
e. \( 2\sqrt{10} + 2\sqrt{6} \) 
PEMBAHASAN : 
\(\begin {eqnarray}\ \frac{\sqrt{8}}{\sqrt{5} - \sqrt{3}} \times \frac {\sqrt{5} + \sqrt{3}} {\sqrt{5} + \sqrt{3}} &=& \frac { \sqrt{40} + \sqrt{24} } { 5 - 3 } \\ &=& \frac { 2\sqrt{10} + 2\sqrt{6}} { 2 } \\ &=& \sqrt {10} + \sqrt{6} \end {eqnarray}\) 
JAWABAN : A 

7. UM UGM TAHUN 2002 
\( \frac{ 5(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})^{3} } { 2\sqrt{2} - \sqrt{3} } \) = … 
a. \( \sqrt{3} - \sqrt{2} \) 
b. \( 3\sqrt{3} - 2\sqrt{2} \) 
c. \( 2\sqrt{2} - 3\sqrt{3} \) 
d. \( 3\sqrt{2} - 2\sqrt{3} \) 
e. \( 4\sqrt{2} - 3\sqrt{3} \) 
PEMBAHASAN : 
\(\begin {eqnarray} \frac{ 5(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})^{3} } { 2\sqrt{2} - \sqrt{3}} &=& \frac { 5(\sqrt{3} + \sqrt{2}) (\sqrt{3} - \sqrt{2}) (\sqrt{3} - \sqrt{2})^{2} } { 2\sqrt{2} - \sqrt{3} } \\ &=& \frac { 5(1)(3 – 2\sqrt{6} + 2) } { 2\sqrt{2} -\sqrt{3} } \\ &=& \frac { 5(5 – 2\sqrt{6}) } { 2\sqrt{2} - 3 } \times \frac { 2\sqrt{2} + 3} { 2\sqrt{2} + 3 } \\ &=& \frac { 5(\sqrt{5} – 2\sqrt{6})(2\sqrt{2} + 3) } { 5 } \\ &=& (5\sqrt{5} – 2\sqrt{6}) (2\sqrt{2} + 3) \\ &=& 10\sqrt{2} + 5\sqrt{3} – 4\sqrt{12} – 2\sqrt{18}\\ &=& 10\sqrt{2} + 5\sqrt{3} – 8\sqrt{3} – 6\sqrt{2} \\ &=& 4\sqrt{2} – 3\sqrt{3} \end {eqnarray}\)
JAWABAN : E 

8. SIMAK UI TAHUN 2009 KODE 921 
Jika \(a = \frac { 2 + \sqrt{3} } { 2 - \sqrt{2} }\) dan \( b = \frac {2 - \sqrt{3}} {2 + \sqrt{3}} \) maka \( a + b = …\) 
a. 0 
b. 1 
c. 8 
d. 10 
e. 14 
PEMBAHASAN : 
\(\begin {eqnarray} a + b &=& \frac { 2 + \sqrt{3} } { 2 - \sqrt{2} } + \frac {2 - \sqrt{3}} {2 + \sqrt{3}} \\ &=& \frac { (2 + \sqrt{3})^{2} + (2 - \sqrt {3})^{2}} { (2 - \sqrt{3})(2 + \sqrt{3}) } \\ &=& \frac { (7 + 4\sqrt{3}) (7 – 4\sqrt{3}) } { 1 } \\ &=& 14 \end {eqnarray}\) 
JAWABAN : E 

9. SPMB 2006 
Jika \(p = (3 + 2\sqrt{2})^{-1} \) dan \( q = (3 – 2\sqrt{2})^{-1} \), maka \( (1 + p)^{-1} + (1 + q)^{-1} =… \) 
a. 0 
b. 1 
c. 8 
d. 10 
e. 14 
PEMBAHASAN :
  • \(\begin {eqnarray} p &=& ( 3 + 2\sqrt{2} )^{-1} \\ &=& \frac { 1 } {3 + 2\sqrt{2} } \\ &=& \frac { 1 } {3 + 2\sqrt{2} } \times \frac { 3 – 2\sqrt{2} } { 3 – 2\sqrt{2} } \\ &=& \frac { 3 – 2\sqrt{2} } { 9 - 8 } \\ &=& 3 – 2\sqrt{2} \end {eqnarray}\)
  • \(\begin {eqnarray} q &=& ( 3 – 2\sqrt{2} )^{-1}\\ &=& \frac { 1 } { 3 – 2\sqrt{2} } \times \frac { 3 + 2\sqrt{2} } { 3 + \sqrt {3} } \\ &=& \frac { 3 + 2\sqrt{2} } { 9 -8 } \\ &=& 3 + 2\sqrt{2} \end {eqnarray}\)
\(\begin {eqnarray} (1 + p )^{-1} + ( 1 + q )^{-1} &=& \frac { 1 } {1 + 3 – 2\sqrt{2} } + \frac { 1 } { 1 + 3 + 2\sqrt{2} } \\ &=& \frac { 1 } {4 – 2\sqrt{2} } + \frac { 1 } { 4 + 2\sqrt{2} } \\ &=& \frac { (4 + 2\sqrt{2}) + (4 – 2\sqrt{2}) } { (4 – 2\sqrt{2} )( 4 + 2\sqrt {2} ) } \\ &=& \frac {8}{8} \\ &=& 1 \end {eqnarray} \) 
JAWABAN : A 

10 SBMTN 2008 KODE 211 
Jika \( \frac { \frac {1}{2} - \frac {1}{\sqrt{5}} } { \frac {1}{2} + \frac {1}{\sqrt{5}} } = a + b\sqrt{5} \), maka \( a + b = …\) 
a. 1 
b. 2 
c. 3 
d. 4 
e. 6 
PEMBAHASAN : 
\(\begin {eqnarray} \frac { \frac {1}{2} - \frac {1}{\sqrt{5}} } { \frac {1}{2} + \frac {1}{\sqrt{5}} } &=& a + b\sqrt{5} \\ \frac { \frac {\sqrt{5} - 2} {2\sqrt{5}} } { \frac {\sqrt{5} + 2}{2\sqrt{5}} } &=& a + b\sqrt{5} \\ \frac { \sqrt{5} - 2}{\sqrt{5} + 2} &=& a + b\sqrt{5} \\ \frac { \sqrt{5} - 2}{\sqrt{5} + 2} \times \frac { \sqrt{5} - 2}{\sqrt{5} - 2} \\ \frac { 5 – 4\sqrt{5} + 4 } { 5 -4 } &=& a + b\sqrt{5} \\ 9 – 4\sqrt{5} &=& a + b\sqrt{5} \end {eqnarray}\)
  • \( a = 9\)
  • \( b = -4\)
  • \( a + b = 5\)
JAWABAN : E