Question

on rationalising the denominator of 1/√3-√2​

Answer 1

Question :-

rationalize the denominator of 1/√3-√2.

Solution :-

$\frac{1}{ \sqrt{3} - \sqrt{2} } \\ \\ \\ multiply \: by \: ( \sqrt{3} + \sqrt{2} ) \: in \: the \: numerator \: and \: denominator \\ \\ \frac{1}{ \sqrt{ 3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \frac{ \sqrt{3} + \sqrt{2} }{ (\sqrt{3} ) {}^{2} - ( \sqrt{2} ) {}^{2} } \\ \\ \frac{ \sqrt{3} + \sqrt{2} }{3 - 2} = (\sqrt{3} + \sqrt{2} ) \: \: \: \: \: \: ans$

Formulae used :-

$1) \: \: \: (x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ \\ 2) \: \: (\sqrt{x}) {}^{2} = x$

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

rationalize the denominator

$\sf\dfrac{1}{\sqrt{3} - \sqrt{2}}$

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

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

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

${\bold{\blue{\boxed{\bf{(a+b)(a-b) = a^2 - b^2 }}}}}$

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

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

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

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

${\bold{\blue{\boxed{\bf{\dag\sqrt 3 + \sqrt 2}}}}}$

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