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 :
- \(\begin {eqnarray} p &=& ( 3 + 2\sqrt{2} )^{-1} \\ &=& \frac { 1 } {3 + 2\sqrt{2} } \\ &=& \frac { 1 } {3 + 2\sqrt{2} } \times \frac { 3 – 2\sqrt{2} } { 3 – 2\sqrt{2} } \\ &=& \frac { 3 – 2\sqrt{2} } { 9 - 8 } \\ &=& 3 – 2\sqrt{2} \end {eqnarray}\)
- \(\begin {eqnarray} q &=& ( 3 – 2\sqrt{2} )^{-1}\\ &=& \frac { 1 } { 3 – 2\sqrt{2} } \times \frac { 3 + 2\sqrt{2} } { 3 + \sqrt {3} } \\ &=& \frac { 3 + 2\sqrt{2} } { 9 -8 } \\ &=& 3 + 2\sqrt{2} \end {eqnarray}\)
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 :
\(\begin {eqnarray}
\frac { \frac {1}{2} - \frac {1}{\sqrt{5}} } { \frac {1}{2} + \frac {1}{\sqrt{5}} } &=& a + b\sqrt{5} \\
\frac { \frac {\sqrt{5} - 2} {2\sqrt{5}} } { \frac {\sqrt{5} + 2}{2\sqrt{5}} } &=& a + b\sqrt{5} \\
\frac { \sqrt{5} - 2}{\sqrt{5} + 2} &=& a + b\sqrt{5} \\
\frac { \sqrt{5} - 2}{\sqrt{5} + 2} \times \frac { \sqrt{5} - 2}{\sqrt{5} - 2} \\
\frac { 5 – 4\sqrt{5} + 4 } { 5 -4 } &=& a + b\sqrt{5} \\
9 – 4\sqrt{5} &=& a + b\sqrt{5}
\end {eqnarray}\)
- \( a = 9\)
- \( b = -4\)
- \( a + b = 5\)
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