Question

Prove that:
$\frac{1}{3 - \sqrt{8} } - \frac{1}{ \sqrt{8 } - \sqrt{7} } + \frac{1}{ \sqrt{7} - \sqrt{6} } - \frac{1}{ \sqrt{6} - \sqrt{5} } + \frac{1}{ \sqrt{5} - 2}$=5
Give step by step explaination.​

Answer 1

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

Consider

$\rm \: \dfrac{1}{3 - \sqrt{8} }$

So, on rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{3 - \sqrt{8} } \times \dfrac{3 + \sqrt{8} }{3 + \sqrt{8} }$

We know,

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

So, using this identity, we get

$\rm \: = \: \dfrac{3 + \sqrt{8} }{ {3}^{2} - {( \sqrt{8}) }^{2} }$

$\rm \: = \: \dfrac{3 + \sqrt{8} }{9 - 8}$

$\rm \: = \: \dfrac{3 + \sqrt{8} }{1}$

$\rm \: = \:3 + \sqrt{8}$

Now, Consider

$\rm \: \dfrac{1}{ \sqrt{8} - \sqrt{7} }$

So, on rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{ \sqrt{8} - \sqrt{7} } \times \dfrac{ \sqrt{8} + \sqrt{7} }{ \sqrt{8} + \sqrt{7} }$

$\rm \: = \: \dfrac{ \sqrt{8} + \sqrt{7} }{ {( \sqrt{8} )}^{2} - {( \sqrt{7} )}^{2} }$

$\rm \: = \: \dfrac{ \sqrt{8} + \sqrt{7} }{8 - 7}$

$\rm \: = \: \dfrac{ \sqrt{8} + \sqrt{7} }{1}$

$\rm \: = \: \sqrt{8} + \sqrt{7}$

Now, Consider

$\rm \: \dfrac{1}{ \sqrt{7} - \sqrt{6} }$

So, on rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{ \sqrt{7} - \sqrt{6} } \times \dfrac{ \sqrt{7} + \sqrt{6} }{ \sqrt{7} + \sqrt{6} }$

$\rm \: = \: \dfrac{ \sqrt{7} + \sqrt{6} }{ {( \sqrt{7} )}^{2} - {( \sqrt{6} )}^{2} }$

$\rm \: = \: \dfrac{ \sqrt{7} + \sqrt{6} }{7 - 6}$

$\rm \: = \: \dfrac{ \sqrt{7} + \sqrt{6} }{1}$

$\rm \: = \: \sqrt{7} + \sqrt{6}$

Now, Consider

$\rm \: \dfrac{1}{ \sqrt{6} - \sqrt{5} }$

So, on rationalizing the denominator, we get

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

$\rm \: = \: \dfrac{ \sqrt{6} + \sqrt{5} }{ {( \sqrt{6} )}^{2} - {( \sqrt{5} )}^{2} }$

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

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

$\rm \: = \: \sqrt{6} + \sqrt{5}$

Now, Consider

$\rm \: \dfrac{1}{ \sqrt{5} - 2}$

So, on rationalizing the denominator, we get

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

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

$\rm \: = \: \dfrac{ \sqrt{5} + 2 }{ 5 - 4 }$

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

$\rm \: = \: \sqrt{5} + 2$

Now, Consider the given expression, we have

$\rm \: \dfrac{1}{3 - \sqrt{8} } - \dfrac{1}{ \sqrt{8 } - \sqrt{7} } + \dfrac{1}{ \sqrt{7} - \sqrt{6} } - \dfrac{1}{ \sqrt{6} - \sqrt{5} } + \dfrac{1}{ \sqrt{5} - 2}$

$\rm \: = \: (3 + \sqrt{8}) - ( \sqrt{8} + \sqrt{7})+( \sqrt{7} + \sqrt{6}) - ( \sqrt{6} + \sqrt{5}) + ( \sqrt{5} + 2)$

$\rm \: = \:3+ \cancel{\sqrt{8}}- \cancel{\sqrt{8}} - \cancel{\sqrt{7}} + \cancel{\sqrt{7}} + \cancel{\sqrt{6}} - \cancel{\sqrt{6}} - \cancel{\sqrt{5}} + \sqrt{5} + 2$

$\rm \: = \: 3 + 2$

$\rm \: = \: 5$

Hence,

$\boxed{\tt{ \frac{1}{3 - \sqrt{8} } - \frac{1}{ \sqrt{8 } - \sqrt{7} } + \frac{1}{ \sqrt{7} - \sqrt{6} } - \frac{1}{ \sqrt{6} - \sqrt{5} } + \frac{1}{ \sqrt{5} - 2} = 5 \: }}$

Answer 2

Question:-

Prove that:

$\sf \large\frac{1}{3 - \sqrt{8} } - \frac{1}{ \sqrt{8 } - \sqrt{7} } + \frac{1}{ \sqrt{7} - \sqrt{6} } - \frac{1}{ \sqrt{6} - \sqrt{5} } + \frac{1}{ \sqrt{5} - 2} =5.$

Solution:-

L.HS.

$\sf \large \frac{1}{3 - \sqrt{8} } - \frac{1}{ \sqrt{8} - \sqrt{7} } + \frac{1}{ \sqrt{7} + 6 } - \frac{1}{ \sqrt{6} - \sqrt{5} } + \frac{1}{ \sqrt{5} - 2}$

$\sf \large = \frac{3 + \sqrt{8} }{(3 - \sqrt{8})(3 + \sqrt{8} )} - \frac{ \sqrt{8} + \sqrt{7} }{( \sqrt{8} - \sqrt{7} )( \sqrt{8} + \sqrt{7}) } + \frac{ \sqrt{7} + \sqrt{6} }{( \sqrt{7} - \sqrt{6} ) (\sqrt{7} + \sqrt{6} )} - \frac{ \sqrt{6} + \sqrt{5} }{( \sqrt{6} - \sqrt{5} )( \sqrt{6} + \sqrt{5} ) } + \frac{ \sqrt{5} + 2}{( \sqrt{5} - 2)( \sqrt{5} + 2)}$

$\sf \large = \frac{3 + \sqrt{8} }{1} - \frac{ \sqrt{8} + \sqrt{7} }{1} + \frac{ \sqrt{7} + \sqrt{6} }{1} - \frac{ \sqrt{6} + \sqrt{5} }{1} + \frac{ \sqrt{5} + 2}{1}$

$\sf \large = 3 {{ \cancel{+ \sqrt{8}}}} {{ \cancel{ - \sqrt{8} }}}{{ \cancel{ - \sqrt{7}}}}{{ \cancel{ + \sqrt{7} }}}{{ \cancel{ + \sqrt{6} }}}{{ \cancel{ - \sqrt{6} }}}{{ \cancel{ - \sqrt{5} }}}{{ \cancel{ + \sqrt{5} }}} + 2$

$\sf \large = 3 + 2$

$\sf \large = 5.$

Answer:-

${ \boxed{ \sf \large \red{\frac{1}{3 - \sqrt{8} } - \frac{1}{ \sqrt{8 } - \sqrt{7} } + \frac{1}{ \sqrt{7} - \sqrt{6} } - \frac{1}{ \sqrt{6} - \sqrt{5} } + \frac{1}{ \sqrt{5} - 2}=5}}}$

${ \boxed{ \sf \large \blue{Hence, Proved \: L.HS. = R.HS.}}}$

Hope you have satisfied. ⚘