Question

please Rationalise the denominator​

Answer 1

Answer :-

$\implies\sf \dfrac{1}{\sqrt3 - \sqrt2 + 1}$

$\implies\sf \dfrac{1}{\sqrt3 - \sqrt2 + 1} \times \dfrac{\sqrt3 - \sqrt2 - 1}{\sqrt3 - \sqrt2 - 1}$

$\implies\sf \dfrac{\sqrt3 - \sqrt2 - 1}{(\sqrt3 - \sqrt2 + 1)({\sqrt3 - \sqrt2 - 1}) }$

• (a + b)(a - b) = a² - b²

$\implies\sf \dfrac{\sqrt3 - \sqrt2 - 1}{(\sqrt3 - \sqrt2)^2 - (1 )^2}$

• (a - b)² = a² - 2ab + b²

$\implies\sf \dfrac{\sqrt3 - \sqrt2 - 1}{(\sqrt3 - \sqrt2)^2 - (1 )^2}$

$\implies\sf \dfrac{\sqrt3 - \sqrt2 - 1}{3 + 2 - 2\sqrt6 - 1}$

$\implies\sf \dfrac{\sqrt3 - \sqrt2 - 1}{4 - 2\sqrt6}$

$\implies\sf \dfrac{\sqrt3 - \sqrt2 - 1}{4 - 2\sqrt6} \times \dfrac{4 + 2\sqrt6}{4 + 2\sqrt6}$

$\implies\sf \dfrac{(\sqrt3 - \sqrt2 - 1)(4 + 2\sqrt6) }{(4 -2\sqrt6)(4+2\sqrt6)}$

$\implies\sf \dfrac{4 (\sqrt3 - \sqrt2 - 1) + 2\sqrt6(\sqrt3 - \sqrt2 - 1)}{4^2 - 4\times 6}$

$\implies\sf \dfrac{4\sqrt3 - 4\sqrt2 - 4 + 6\sqrt2 - 4\sqrt3 + 2\sqrt6}{16 - 24}$

$\implies\sf \dfrac{2\sqrt2 - 4 - 2\sqrt6}{-8}$

$\implies\sf \dfrac{-2(2 - \sqrt2 + \sqrt6)}{-8}$

$\implies\boxed{\sf \dfrac{ 2 + \sqrt6 - \sqrt2}{4}}$