Rationalise the denominators of the following:
(i) $ \frac{1}{\sqrt{7}} $
(ii) $ \frac{1}{\sqrt{7}-\sqrt{6}} $
(iii) $ \frac{1}{\sqrt{5}+\sqrt{2}} $
(iv) $ \frac{1}{\sqrt{7}-2} $

To do:

We have to rationalise the denominators of the given expressions.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Therefore,

(i) $\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}$

$=\frac{\sqrt{7}}{(\sqrt{7})^2}$

$=\frac{\sqrt{7}}{7}$.

(ii) $\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\times\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$

$=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2-(\sqrt{6})^2}$

$=\frac{\sqrt{7}+\sqrt{6}}{7-6}$

$=\sqrt{7}+\sqrt{6}$.

(iii) $\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{\sqrt{5}+\sqrt{2}}\times\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

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

$=\frac{\sqrt{5}-\sqrt{2}}{5-2}$

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

(iv) $\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times\frac{\sqrt{7}+2}{\sqrt{7}+2}$

$=\frac{\sqrt{7}+2}{(\sqrt{7})^2-(2)^2}$

$=\frac{\sqrt{7}+2}{7-4}$

$=\frac{\sqrt{7}+2}{3}$.

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Updated on: 10-Oct-2022

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