Question

Help me to solve this problem

Answer 1

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Answer 2

Solution :

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Given :

To find the value of :

$\frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3}+ \sqrt{2} } - \frac{ \sqrt{3}+ \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

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By cross-multiplication,

We get,.

⇒ $\frac{( \sqrt{3} - \sqrt{2} )^2 - ( \sqrt{3} + \sqrt{2} )^2}{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} )}$

We know that,

(a - b)(a + b) = a² - b²

⇒ $\frac{ (\sqrt{3})^2 - 2( \sqrt{3})( \sqrt{2}) + ( \sqrt{2})^2 - [( \sqrt{3})^2 + 2( \sqrt{3})( \sqrt{2}) + (\sqrt{2})^2 }{( \sqrt{3})^2 - ( \sqrt{2})^2}$

⇒ $\frac{(3 + 2 - 2 \sqrt{6} ) - (3 + 2 + 2 \sqrt{6})}{3 - 2}$

⇒ $\frac{(5 - 2 \sqrt{6}) - (5 + 2\sqrt{6} )}{1}$

⇒ $5 - 2 \sqrt{6} - 5 - 2 \sqrt{6}$

⇒ $-2 \sqrt{6} - 2\sqrt{6}$

⇒ $-4 \sqrt{6}$

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                                        Hope it Helps !!

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