Show that $ \frac{a \sqrt{b}-b \sqrt{a}}{a \sqrt{b}+b \sqrt{a}}=\frac{1}{a-b}(a+b-2 \sqrt{a b}) $

To do :

We have to show that \( \frac{a \sqrt{b}-b \sqrt{a}}{a \sqrt{b}+b \sqrt{a}}=\frac{1}{a-b}(a+b-2 \sqrt{a b}) \).

Solution :

LHS $=\frac{a \sqrt{b}-b \sqrt{a}}{a \sqrt{b}+b \sqrt{a}}$

To factorize the denominator, multiply and divide by $a \sqrt{b}-b \sqrt{a}$.

LHS

$\frac{a \sqrt{b}-b \sqrt{a}}{a \sqrt{b}+b \sqrt{a}} = \frac{(a \sqrt{b}-b \sqrt{a}) \times (a \sqrt{b}-b \sqrt{a})}{(a \sqrt{b}+b \sqrt{a}) \times (a \sqrt{b}-b \sqrt{a})}$

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

$ = \frac{a^2b + b^2a - 2 \times a\sqrt{b} \times b\sqrt{a}}{a^2b-b^2a}$

$ = \frac{ab(a+b-2\sqrt{ab})}{ab(a-b)}$

$ =\frac{1}{a-b}(a+b-2 \sqrt{a b})$

$=$ RHS

Hence proved.

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Updated on: 10-Oct-2022

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