Question

rationalise the denominator. 1/√6+√5-√11

Answer 1

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Answer 2

Answer:

$\frac{1}{60}(6\sqrt{5}+5\sqrt{6}+\sqrt{330})$

Step-by-step explanation:

Here, the given expression is,

$\frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}$

For rationalizing the denominator, multiply both numerator and denominator by √6 + √5 + √11,

$=\frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}\times \frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{\sqrt{6}+\sqrt{5}+\sqrt{11}}$

$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6}+\sqrt{5})^2-(\sqrt{11})^2}$

$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{6+5+2\times\sqrt{6}\times \sqrt{5}-11}$

$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{2\sqrt{30}}$

Again for rationalizing the denominator, multiply both numerator and denominator by √30,

$=\frac{\sqrt{180}+\sqrt{150}+\sqrt{330}}{60}$

$=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{60}$

$=\frac{1}{60}(6\sqrt{5}+5\sqrt{6}+\sqrt{330})$