Question

pls solve the above question ↑
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Answer 1

Question :

$\sf \small \frac{1}{2 + \sqrt{5} } + \frac{1}{ \sqrt{5} + \sqrt{6} } + \frac{1}{ \sqrt{6} + \sqrt{7} } + \frac{1}{ \sqrt{7} + \sqrt{8} } + \frac{1}{ \sqrt{8} + \sqrt{9} }$

Solution :

$\sf \small \frac{1}{2 + \sqrt{5} } + \frac{1}{ \sqrt{5} + \sqrt{6} } + \frac{1}{ \sqrt{6} + \sqrt{7} } + \frac{1}{ \sqrt{7} + \sqrt{8} } + \frac{1}{ \sqrt{8} + \sqrt{9} }$

Now we have to rationalize the denominator

$\sf \tiny \frac{1(2 - \sqrt{5} )}{(2+ \sqrt{5})(2 - \sqrt{5})} + \frac{1( \sqrt{5} - \sqrt{6} )}{( \sqrt{5} + \sqrt{6} )( \sqrt{5} - \sqrt{6})} + \frac{1( \sqrt{6} - \sqrt{7} )}{( \sqrt{6} + \sqrt{7})( \sqrt{6} - \sqrt{7}) } + \frac{1( \sqrt{7} - \sqrt{8}) }{( \sqrt{7} + \sqrt{8})( \sqrt{7} - \sqrt{8} ) } + \frac{1( \sqrt{8} - \sqrt{9} )}{( \sqrt{8} + \sqrt{9} )( \sqrt{8} - \sqrt{9} )}$

Now we need to simplify the denominator using (a + b)(a - b) → a² – b²

√9 can be written as 3 as 3² = 9

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

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

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

Now taking -1 as common we get

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

All the no.s in the numerator will get cancelled because of opposite signs

$\sf \small \frac{2 - 3 }{ - 1}$

$\sf \small \frac{ - 1}{ - 1}$

$1$

So the result is 1