The correct option is B 11 2√(30) √(6) √(5)/√(6)+ √(5)×√(6) √(5)/√(6) √(5) =(√(6) √(5))^2/(√(6))^2 (√(5))^2 =(√(6))^2 2(√(6))(√(5))+(√(5))^2/6 5 = 6 2 √ ...

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Monomial and Binomial Surds

√6-√5/√6+√5=

Question

$\frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}$ = ________.

$4 \sqrt{30}$

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$11 - 2 \sqrt{30}$

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$11 - \sqrt{30}$

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Solution

The correct option is C
$11 - 2 \sqrt{30}$

$\frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}}$

= $\frac{(\sqrt{6} - \sqrt{5})^{2}}{(\sqrt{6})^{2} - (\sqrt{5})^{2}}$

= $\frac{(\sqrt{6})^{2} - 2 (\sqrt{6}) (\sqrt{5}) + (\sqrt{5})^{2}}{6 - 5}$

= $\frac{6 - 2 \sqrt{30} + 5}{1}$

= $11 - 2 \sqrt{30}$

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Similar questions

Q. If

\frac{4^{5} + 4^{5} + 4^{5} + 4^{5}}{2^{5} + 2^{5}} \times \frac{6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5}}{3^{5} + 3^{5} + 3^{5}} = 8^{n}

, then n is:

Q. Let

N = \frac{4^{5} + 4^{5} + 4^{5} + 4^{5}}{3^{5} + 3^{5} + 3^{5}} . \frac{6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5}}{2^{5} + 2^{5}}

, then the value of

{log}_{2} N =

Let N = $\frac{4^{5} + 4^{5} + 4^{5} + 4^{5}}{3^{5} + 3^{5} + 3^{5}} \cdot \frac{6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5}}{2^{5} + 2^{5}}$ then the value of ${log}_{2} N$ =

{[\frac{5}{- 6}]}^{4} \div [\frac{5}{- 6}] X = {{[\frac{5}{- 6}]}^{2}}^{3}

Q. Factorise:

25 a^{2} - 36 b^{2}

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√6−√5√6+√5 = ________.

The correct option is C 11−2√30 √6−√5√6+√5×√6−√5√6−√5 =(√6−√5)2(√6)2−(√5)2 =(√6)2−2(√6)(√5)+(√5)26−5 = 6−2√30+51 = 11−2√30

$\frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}$ = ________.