Question

Rationalise The Denominator $\frac{1}{1+\sqrt{2}+\sqrt{3} }$

Answer 1

$\large\underline{\sf{Solution-}}$

Given irrational number is

$\rm :\longmapsto\:\dfrac{1}{1 + \sqrt{2} + \sqrt{3} }$

can be rewritten as

$\rm \: = \:\dfrac{1}{ \sqrt{3} + \sqrt{2} + 1}$

So, on rationalizing the denominator, we get

$\rm \: = \:\dfrac{1}{ (\sqrt{3} + \sqrt{2})+ 1} \times \dfrac{( \sqrt{3} + \sqrt{2} ) - 1}{( \sqrt{3} + \sqrt{2) - 1} }$

We know,

$\boxed{ \tt{ \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \: = \:\dfrac{ \sqrt{3} + \sqrt{2} - 1 }{ {( \sqrt{3} + \sqrt{2} )}^{2} - {1}^{2} }$

$\rm \: = \:\dfrac{ \sqrt{3} + \sqrt{2} - 1}{3 + 2 + 2 \times \sqrt{2} \times \sqrt{3} - 1}$

$\rm \: = \:\dfrac{ \sqrt{3} + \sqrt{2} - 1}{5 + 2 \sqrt{6} - 1}$

$\rm \: = \:\dfrac{ \sqrt{3} + \sqrt{2} - 1}{4 + 2 \sqrt{6}}$

$\rm \: = \:\dfrac{ \sqrt{3} + \sqrt{2} - 1}{4 + 2 \sqrt{6}} \times \dfrac{4 - 2 \sqrt{6} }{4 - 2 \sqrt{6} }$

$\rm \: = \:\dfrac{4 \sqrt{3} + 4 \sqrt{2} - 4 - 2 \sqrt{18} - 2 \sqrt{12} + 2 \sqrt{6} }{ {4}^{2} - {(2 \sqrt{6} )}^{2} }$

$\rm \: = \:\dfrac{4 \sqrt{3} + 4 \sqrt{2} - 4 - 6\sqrt{2} - 4\sqrt{3} + 2 \sqrt{6} }{16 - 24 }$

$\rm \: = \:\dfrac{ - 2 \sqrt{2} - 4 + 2 \sqrt{6} }{ - 8}$

$\rm \: = \:\dfrac{ - 2 (\sqrt{2} + 2 - \sqrt{6} )}{ - 8}$

$\rm \: = \:\dfrac{\sqrt{2} + 2 - \sqrt{6} }{4}$

Alter method

$\rm :\longmapsto\:\dfrac{1}{1 + \sqrt{2} + \sqrt{3} }$

$\rm \: = \:\dfrac{1}{ \sqrt{3} + \sqrt{2} + 1}$

$\rm \: = \:\dfrac{1}{ \sqrt{3} + (\sqrt{2} + 1)} \times \dfrac{ \sqrt{3} - ( \sqrt{2} + 1)}{ \sqrt{3} - ( \sqrt{2} + 1)}$

$\rm \: = \:\dfrac{ \sqrt{3} - \sqrt{2} - 1}{ {( \sqrt{3}) }^{2} - {( \sqrt{2} - 1)}^{2} }$

$\rm \: = \:\dfrac{ \sqrt{3} - \sqrt{2} - 1}{ {3 - (2 + 1 + 2 \sqrt{2} )} }$

$\rm \: = \:\dfrac{ \sqrt{3} - \sqrt{2} - 1}{ {3 - 2 - 1 - 2 \sqrt{2}} }$

$\rm \: = \:\dfrac{ \sqrt{3} - \sqrt{2} - 1}{ { - 2 \sqrt{2}} }$

$\rm \: = \:\dfrac{ \sqrt{3} - \sqrt{2} - 1}{ { - 2 \sqrt{2}} } \times \dfrac{ \sqrt{2} }{ \sqrt{2} }$

$\rm \: = \:\dfrac{ \sqrt{6} - 2 - \sqrt{2} }{ - 4}$

$\rm \: = \:\dfrac{\sqrt{2} + 2 - \sqrt{6} }{4}$

Hence,

$\bf\implies \:\boxed{ \tt{ \: \frac{1}{ \sqrt{3} + \sqrt{2} + 1 } = \:\dfrac{\sqrt{2} + 2 - \sqrt{6} }{4}}}$