Question

prove that (1)/(3+sqrt(7))+(1)/(sqrt(7)+sqrt(5))+(1)/(sqrt(5)+sqrt(3))+(1)/(sqrt(3)+1)=1

Answer 1

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

Given expression is

$\rm :\longmapsto\:\dfrac{1}{3 + \sqrt{7} } + \dfrac{1}{ \sqrt{7} + \sqrt{5} } + \dfrac{1}{ \sqrt{5} + \sqrt{3} } + \dfrac{1}{ \sqrt{3} + 1}$

Consider,

$\red{\rm :\longmapsto\:\dfrac{1}{3 + \sqrt{7} }}$

On rationalizing the denominator, we get

$\rm \: = \: \: \dfrac{1}{3 + \sqrt{7} } \times \dfrac{3 - \sqrt{7} }{3 - \sqrt{7} }$

$\rm \: = \: \: \dfrac{3 - \sqrt{7} }{ {(3)}^{2} - {( \sqrt{7} )}^{2} }$

$\red{\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \: = \: \: \dfrac{3 - \sqrt{7} }{9 - 7}$

$\rm \: = \: \: \dfrac{3 - \sqrt{7} }{2}$

$\red{\bf :\longmapsto\:\dfrac{1}{3 + \sqrt{7} } = \: \: \dfrac{3 - \sqrt{7} }{2} - - - (1)}$

Consider,

$\green{\rm :\longmapsto\:\dfrac{1}{ \sqrt{7} + \sqrt{5} }}$

On rationalizing the denominator, we get

$\rm \: = \: \: \dfrac{1}{ \sqrt{7} + \sqrt{5} } \times \dfrac{ \sqrt{7} - \sqrt{5} }{ \sqrt{7} - \sqrt{5} }$

$\rm \: = \: \: \dfrac{ \sqrt{7} - \sqrt{5} }{ {( \sqrt{7} )}^{2} - {( \sqrt{5} )}^{2} }$

$\rm \: = \: \: \dfrac{ \sqrt{7} - \sqrt{5} }{7 - 5}$

$\rm \: = \: \: \dfrac{ \sqrt{7} - \sqrt{5} }{2}$

$\green{\bf :\longmapsto\:\dfrac{1}{ \sqrt{7} - \sqrt{5} } = \: \: \dfrac{ \sqrt{7} - \sqrt{5} }{2} - - - (2)}$

Consider,

$\blue{\rm :\longmapsto\:\dfrac{1}{ \sqrt{5} + \sqrt{3} }}$

On rationalizing the denominator, we get

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

$\rm \: = \: \: \dfrac{ \sqrt{5} - \sqrt{3} }{ {( \sqrt{5} )}^{2} - {( \sqrt{3}) }^{2} }$

$\rm \: = \: \: \dfrac{ \sqrt{5} - \sqrt{3} }{5 - 3}$

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

$\blue{\bf :\longmapsto\:\dfrac{1}{ \sqrt{5} + \sqrt{3} } = \: \: \dfrac{ \sqrt{5} - \sqrt{3} }{2} - - - (3)}$

Consider,

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

On rationalizing the denominator, we get

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

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

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

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

$\purple{\bf :\longmapsto\:\dfrac{1}{ \sqrt{3} + 1} = \: \: \dfrac{ \sqrt{3} - 1}{2} - - - (4)}$

Now, Consider

$\pink{\rm :\longmapsto\:\dfrac{1}{3 + \sqrt{7} } + \dfrac{1}{ \sqrt{7} + \sqrt{5} } + \dfrac{1}{ \sqrt{5} + \sqrt{3} } + \dfrac{1}{ \sqrt{3} + 1} }$

On substituting the values evaluated in above steps, we get

$\rm \: = \: \: \dfrac{3 - \sqrt{7} }{2} + \dfrac{ \sqrt{7} - \sqrt{5} }{2} + \dfrac{ \sqrt{5} - \sqrt{3} }{2} + \dfrac{ \sqrt{3} - 1}{2}$

$\rm \: = \: \: \dfrac{3 - \sqrt{7} + \sqrt{7} - \sqrt{5} + \sqrt{5} - \sqrt{3} + \sqrt{3} - 1 }{2}$

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

$\rm \: = \: \: \dfrac{2}{2}$

$\rm \: = \: \: 1$

Hence, Proved

Answer 2

Answer:

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