Question

rationalise the denominator​

Answer 1

$\underline{\underline{\sf{\maltese\:\: Solution }}}$

$\sf(a) \: \: \: \dfrac{1}{ \sqrt{3} - \sqrt{2} }$

On rationalising,

$\sf \implies \dfrac{1}{ \sqrt{3} - \sqrt{2} } \times \dfrac{\sqrt{3} + \sqrt{2} }{\sqrt{3} + \sqrt{2} }$

Combine the fractions,

$\sf \implies \dfrac{1(\sqrt{3} + \sqrt{2}) }{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2}) }$

We know that,

$\sf \implies (a - b)(a + b) = (a)^{2} - (b)^{2}$

So,

$\sf \implies \dfrac{1(\sqrt{3} + \sqrt{2}) }{(\sqrt{3})^{2} - (\sqrt{2}) ^{2} }$

$\sf \implies \dfrac{1(\sqrt{3} + \sqrt{2}) }{3 - 2 }$

$\sf \implies \dfrac{1(\sqrt{3} + \sqrt{2}) }{1 }$

$\sf \implies ( \sqrt{3} + \sqrt{2})$

Hence,

On rationalising we got,

$\implies \bf (\sqrt{3} + \sqrt{2})$

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$\sf(b) \: \: \: \dfrac{1}{ 2 \sqrt{5} + \sqrt{3} }$

On rationalising,

$\sf \implies\dfrac{1}{ 2 \sqrt{5} + \sqrt{3} } \times \dfrac{2 \sqrt{5} - \sqrt{3}}{2 \sqrt{5} - \sqrt{3}}$

Combine the fractions,

$\sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{(2 \sqrt{5} + \sqrt{3})(2 \sqrt{5} - \sqrt{3})}$

We know that,

$\sf \implies (a + b)(a - b) = (a)^{2} - (b)^{2}$

So,

$\sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{(2 \sqrt{5} )^{2} - (\sqrt{3})^{2} }$

$\sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{20 - 3 }$

$\sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{17 }$

$\sf \implies \dfrac{2 \sqrt{5} - \sqrt{3}}{17 }$

Hence,

On rationalising we got,

$\bf \implies\dfrac{2 \sqrt{5} - \sqrt{3}}{17 }$