Question

A) 4 b) 2

explain with proof

Answer 1

Here is how,

on rationalising,

$\frac{1}{1 + \sqrt{2} } \\ = > \frac{1(1 - \sqrt{2}) }{(1 + \sqrt{2} )(1 - \sqrt{2)} } \\ = > \frac{1 - \sqrt{2} }{1 - 2} \\ = > \frac{1 - \sqrt{2} }{ - 1} \\ \\ = > \sqrt{2 } - 1$
Similarly, if you rationalize others, we can see,
$\frac{1}{ \sqrt{2} + \sqrt{3} } \\ \\ = \sqrt{3} - \sqrt{2} \\ \\ \frac{1}{ \sqrt{3} + \sqrt{4} } = \sqrt{4} - \sqrt{3} \\ = 2 - \sqrt{3} \\ \\ \frac{1}{ \sqrt{4} + \sqrt{5} } = \sqrt{5 } - \sqrt{4} \\ = > \sqrt{5 } - 2 \\ \\ \\ \frac{1}{ \sqrt{5} + \sqrt{6} } = \sqrt{6} - \sqrt{5 } \\ \\ \frac{1}{ \sqrt{6} + \sqrt{7} } = \sqrt{7} - \sqrt{6} \\ \\ \frac{1}{ \sqrt{7} + \sqrt{8} } = \sqrt{8} - \sqrt{7} \\ \\ \frac{1}{ \sqrt{8} } + \sqrt{9} = \sqrt{9} - \sqrt{8} \\ = > 3 - \sqrt{8}$
now if we add them then,

$\sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} + \sqrt{5} - 2 + \sqrt{6} - \sqrt{5} + \sqrt{7} - \sqrt{6} + \sqrt{8} - \sqrt{7} + 3 - \sqrt{8} \\ \\ = > - 1 + 3 \\ \\ = > 2$
hence, 2 is your answer


Hope it helps dear friend ☺️✌️