Question

please help mates I will definitely mark as brainllist and thank and one who has a poor answer will be deleted.

please help
50 POINT NEED STEP- BY- STEP EXPLAINATION.​

Answer 1

hiii....

it can be solved directly by taking product All denominator as LCM but will be a bit complicated.....

so,.we will,use,the rationalization method

LHS=(3+√8)/{(3-√8)(3+√8) -(√8+√7)/{(√8+√7)(√8-√7)} +(√7+√6)/{(√7-√6)(√7+√6) -(√6+√5)/{(√6-√5)(√6+√5)} +(√5+2)/{(√5-2)(√5+2)}

using the identity ( a+b)(a-b)=a^2-b^2,,simplify the denominator

=(3+√8)(9-8) -(√8+√7)/(8-7)+(√7+√6)/(7-6) -(√6+√5)/(6-5) +(√5+2)/(5-4)

=3+√8-√8-√7+√7+√6-√6-√5+√5+2

=3+2

=5=RHS

Answer 2

Answer:

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

By the method of Rationalization or by Conjugation,

Take each terms seperately and do..

Here I am doing this by conjugation method,

$\frac{1}{3-\sqrt{8} } =\frac{1}{3-\sqrt{8} } × \frac{3+\sqrt{8} }{ 3+\sqrt{8} } =3+\sqrt{8}$

Then similarly,

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

Now taking the LHS as,

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

which is equal to RHS..

Therefore it is proved...

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