Question

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

Answer 1

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

Given expression is

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

Now, Consider

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

On rationalizing the denominator, we get

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

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

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

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

$\rm \:  =  \: 3 + \sqrt{8}$

Now, Consider

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

On rationalizing the denominator, we get

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

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

$\rm \:  =  \: \sqrt{8} + \sqrt{7}$

Now, Consider

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

On rationalizing the denominator, we get

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

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

$\rm \:  =  \: \sqrt{7} + \sqrt{6}$

Now, Consider

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

On rationalizing the denominator, we get

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

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

$\rm \:  =  \: \sqrt{6} + \sqrt{5}$

Now, Consider

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

On rationalizing the denominator, we get

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

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

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

$\rm \:  =  \: \sqrt{5} + 2$

Hence,

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

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

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

$\rm \:  =  \: 3 + 2$

$\rm \:  =  \: 5$

Hence,

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

Answer 2

Answer:

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

Given expression is

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

Now, Consider

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

On rationalizing the denominator, we get

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

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

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

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

$\rm \:  =  \: 3 + \sqrt{8}$

Now, Consider

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

On rationalizing the denominator, we get

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

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

$\rm \:  =  \: \sqrt{8} + \sqrt{7}$

Now, Consider

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

On rationalizing the denominator, we get

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

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

$\rm \:  =  \: \sqrt{7} + \sqrt{6}$

Now, Consider

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

On rationalizing the denominator, we get

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

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

$\rm \:  =  \: \sqrt{6} + \sqrt{5}$

Now, Consider

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

On rationalizing the denominator, we get

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

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

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

$\rm \:  =  \: \sqrt{5} + 2$

Hence,

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

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

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

$\rm \:  =  \: 3 + 2$

$\rm \:  =  \: 5$

Hence,

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