Question

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Answer 1

[tex]\mathfrak{\huge{Answer:-}} \\ \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 \\ \frac{1}{3 - \sqrt{8} } \times \frac{3 + \sqrt{8} }{3 + \sqrt{8} } - \frac{1}{ \sqrt{8} - \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} - 2} \times \frac{ \sqrt{5} + 2}{ \sqrt{5} + 2 } = 5 \\ \frac{1(3 + \sqrt{8}) }{(3 - \sqrt{8})(3 + \sqrt{8})} - \frac{1( \sqrt{8} + \sqrt{7})}{( \sqrt{8} - \sqrt{7})( \sqrt{8} + \sqrt{7})} + \frac{1( \sqrt{7} + \sqrt{6})}{( \sqrt{7} - \sqrt{6})( \sqrt{7} + \sqrt{6})} - \frac{1( \sqrt{6} + \sqrt{5})}{( \sqrt{6} - \sqrt{5})( \sqrt{6} + \sqrt{5})} + \frac{1( \sqrt{5} + 2)}{( \sqrt{5} - 2)( \sqrt{5} + 2) } = 5 \\ \frac{3 + \sqrt{8} }{ {(3)}^{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} + 2}{ {( \sqrt{5} )}^{2} - {(2)}^{2} } = 5 \\ \frac{3 + \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} + 2 }{5 - 4} = 5 \\ \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} = 5 \\ (3 + \sqrt{8} ) - ( \sqrt{8} + \sqrt{7} ) + ( \sqrt{7} + \sqrt{6} ) - ( \sqrt{6} + \sqrt{5} ) + ( \sqrt{5} + 2) = 5 \\ 3 + \sqrt{8} - \sqrt{8} - \sqrt{7} + \sqrt{7} + \sqrt{6} - \sqrt{6} - \sqrt{5} + \sqrt{5} + 2 = 5 \\ 3 + 2 + \sqrt{8} - \sqrt{8} + \sqrt{7} - \sqrt{7} + \sqrt{6} - \sqrt{6} + \sqrt{5} - \sqrt{5} = 5 \\ 5 = 5\\LHS=RHS\\Hence\: proved [/tex]