Question

Rationalise the denominator is 1/(√4 - √2)​

Answer 1

Answer:

$= { \tt{ \frac{1}{( \sqrt{4} - \sqrt{2} )}. \frac{ \sqrt{4} + \sqrt{2} }{ \sqrt{4} + \sqrt{2} } }} \\ \\ = { \tt{ \frac{ \sqrt{4} + \sqrt{2} }{( \sqrt{4}) {}^{2} - {( \sqrt{2}) }^{2} } }} \\ \\ = { \tt{ \frac{ \sqrt{4} + \sqrt{2} }{4 - 2} }} \\ \\ = { \tt{ \frac{2 + \sqrt{2} }{2} } } \\ \\ = { \boxed{ \tt{ \: \: 1 + \frac{ \sqrt{2} }{2} }}}$

Answer 2

Step-by-step explanation:

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

$\sf(a) \: \: \: \dfrac{1}{ \sqrt{4} - \sqrt{2} }$

On rationalising,

$\sf \implies \dfrac{1}{ \sqrt{4} - \sqrt{2} } \times \dfrac{\sqrt{4} + \sqrt{2} }{\sqrt{4} + \sqrt{2} }$

Combine the fractions,

$\sf \implies \dfrac{1(\sqrt{4} + \sqrt{2}) }{(\sqrt{4} - \sqrt{2})(\sqrt{4} + \sqrt{2}) }$

We know that,

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

So,

$\sf \implies \dfrac{1(\sqrt{4} + \sqrt{2}) }{(\sqrt{4})^{2} - (\sqrt{2}) ^{2} }$

$\sf \implies \dfrac{1(\sqrt{4} + \sqrt{2}) }{4 - 2 }$

$\sf \implies \dfrac{1(\sqrt{4} + \sqrt{2}) }{2 }$

$\sf \implies \dfrac{\sqrt{4} + \sqrt{2}}{2}$

Hence,

Hence, On rationalising we got,

$\implies \bf {\dfrac{\sqrt{4} + \sqrt{2}}{2}}$