Question

Show that 1/3√8-1/√8-√7-1/√7-√6-1/√6-√5-1/√5+2 = 5
please give answer fast​

Answer 1

Correct Question :

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

For solving this we need to use the concept of rationalisation

Rationalisation :

• Rationalisation is a process by which radicals in the denominator of an algebraic fraction are eliminated.

• However, except in special cases, the resulting fractions may have huge numerators and denominators, and, therefore, the technique is generally used only in the above elementary cases.

Solution :

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

$\rarr \tiny\frac{1}{3 -\sqrt{8} } - \frac{1( \sqrt{8} + \sqrt{7} ) }{( \sqrt{8} - \sqrt{7})( \sqrt{8} + \sqrt{7} )} + \frac{1( \sqrt{7} + \sqrt{6} )}{( \sqrt{7} - \sqrt{6} )( \sqrt{7} + \sqrt{6} )} - \frac{1( \sqrt{6} + \sqrt{5} )}{( \sqrt{6} - \sqrt{5} )( \sqrt{6} + \sqrt{5}) } + \frac{1( \sqrt{5} + 2) }{( \sqrt{5} - 2)( \sqrt{5} +2) } = 5$

$\rarr\small\frac{3+\sqrt{8}}{(3)² -(\sqrt{8} )^{2} } - \frac{ \sqrt{8} + \sqrt{7} }{ {( \sqrt{8}) }^{2} - ( \sqrt{7} )^{2} } + \frac{ \sqrt{7} + \sqrt{6} }{ {( \sqrt{7} )}^{2} - {( \sqrt{6} )}^{2} } - \frac{ \sqrt{6} + \sqrt{5} }{ {( \sqrt{6}) }^{2} - ( \sqrt{5})^{2} } + \frac{ \sqrt{5} + 2 }{ {( \sqrt{5}) }^{2} - {(2)}^{2} } = 5$

$\rarr \small \frac{ 3+\sqrt{8} }{9-8} - \frac{ \sqrt{8} + \sqrt{7} }{8 - 7} - \frac{ \sqrt{7} + \sqrt{6} }{7 - 6} + \frac{ \sqrt{6} + \sqrt{5} }{6 - 5} + \frac{ \sqrt{5} +2}{5 - 4} = 5$

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

$\rarr \small3 + \sqrt{8} - \sqrt{8} + \sqrt{7} - \sqrt{7} + \sqrt{6} - \sqrt{6} - \sqrt{5} + \sqrt{5} + 2 = 5$

$\rarr \small3 + 2 = 5$

$\rarr \small5 = 5$

Hence Verified !!