Question

plz guys solve this

Answer 1

Onkay, So first what we're gonna do is rationalize each number.

Here we go,

$\frac{1}{2 + \sqrt{5} }$
$\frac{1}{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} }$

Now as we can see the bas is in the form of (a+b)(a-b) which is the same thing as a^2 - b^2
so it becomes
$\frac{2 - \sqrt{5} }{ {2}^{2} - { \sqrt{5} }^{2} } \\ = \frac{2 - \sqrt{5} }{4 - 5} \\ = \frac{2 - \sqrt{5} }{ - 1} \\ = \sqrt{5} - 2$
similarly, others could also be rationalised as shown in the image above.

Therefore,

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

Add all of them together and you'll get

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