Question

$\frac{?}{ \sqrt{128} } = \frac{ \sqrt{162 =} }{?}$
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Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore \frac{1}{( \sqrt{9} - \sqrt{8} ) } - \frac{1}{ (\sqrt{8} - \sqrt{7} )} + \frac{1}{ (\sqrt{7} - \sqrt{6}) } - \frac{1}{ (\sqrt{6} - \sqrt{5}) } + \frac{1}{ \sqrt{5} - \sqrt{4} } =5}}[tex]</p><p></p><p>[tex]$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\bold{for \: finding \: given \: values : } \\ \implies \frac{1}{( \sqrt{9} - \sqrt{8} ) } - \frac{1}{ (\sqrt{8} - \sqrt{7} )} + \frac{1}{ (\sqrt{7} - \sqrt{6}) } - \frac{1}{ (\sqrt{6} - \sqrt{5}) } + \frac{1}{ \sqrt{5} - \sqrt{4} } \\ \\ \bold{rationalise \: denominators : } \\ \implies \frac{1}{( \sqrt{9} - \sqrt{8} ) } \times \frac{ \sqrt{9} + \sqrt{8} }{ (\sqrt{9} + \sqrt{8} ) } - \frac{1}{ (\sqrt{8} \times - \sqrt{7} )} \times \frac{ \sqrt{8 } + \sqrt{7} }{ (\sqrt{8} + \sqrt{7}) } + \frac{1}{ (\sqrt{7} - \sqrt{6}) } \times \frac{ \sqrt{7} + \sqrt{6} }{ (\sqrt{7} + \sqrt{6}) } - \frac{1}{ (\sqrt{6} - \sqrt{5}) } \times \frac{ \sqrt{6} + \sqrt{5} }{( \sqrt{6} + \sqrt{5} ) } + \frac{1}{ (\sqrt{5} - \sqrt{4}) } \times \frac{ \sqrt{5} + \sqrt{4} }{ (\sqrt{5} + \sqrt{4}) } \\ \\ \bold{ ({a} + b)(a - b) = {a}^{2} - {b}^{2} } \\ \\ \implies \frac{ \sqrt{9} + \sqrt{8} }{ ({\sqrt{9}})^{2} - { (\sqrt{8} })^{2} } - \frac{ \sqrt{8} + \sqrt{7} }{ ({ \sqrt{8} })^{2} - {( \sqrt{7} })^{2} } + \frac{ \sqrt{7} + \sqrt{6} }{ { (\sqrt{7} })^{2} - {( \sqrt{6} })^{2} } - \frac{ \sqrt{6} + \sqrt{5} }{ { (\sqrt{6} })^{2} - { (\sqrt{5} })^{2} } + \frac{ \sqrt{5} + \sqrt{4} }{ ({ \sqrt{5} })^{2} - { (\sqrt{4} })^{2} } \\ \\ \implies \frac{ \sqrt{9} + \sqrt{8} }{ 9 - 8} - \frac{ \sqrt{8} + \sqrt{7} }{8 - 7} + \frac{ \sqrt{7} + \sqrt{6} }{7 - 6} - \frac{ \sqrt{6} + \sqrt{5} }{6 - 5} + \frac{ \sqrt{5} + \sqrt{4} }{5 - 4} \\ \\ \implies \sqrt{9} + \cancel{\sqrt{8}} - \cancel{\sqrt{8}} - \cancel{\sqrt{7}} + \cancel{\sqrt{7} } + \cancel{ \sqrt{6} } - \cancel{\sqrt{6} } - \cancel{\sqrt{5}} + \cancel{\sqrt{5}} + \sqrt{4} \\ \\ \implies \sqrt{9} + \sqrt{4} \\ \\ \implies 3 + 2 \\ \\ \bold{\implies 5}$