Question

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Answer 1

$\sf \Large Solution!!$

$\sf \large To\:prove:$

$\sf \to \dfrac{1}{2+\sqrt{3}}+\dfrac{2}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{2-\sqrt{5}}=0$

$\sf Solving\:LHS!!$

Rationalize the denominator.

$\sf \dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}+\dfrac{2}{\sqrt{5}-\sqrt{3}}\times \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{1}{2-\sqrt{5}}\times \dfrac{2+\sqrt{5}}{2+\sqrt{5}}$

$\sf \to \cancel{2}\cancel{-\sqrt{3}}\cancel{+\sqrt{5}}\cancel{+\sqrt{3}}\cancel{-2}\cancel{-\sqrt{5}}$

$\sf \to L.H.S=R.H.S$

Hence, proved.

Answer 2

Step-by-step explanation:

L.H.S.

(✓5-✓3)(2-✓5)+2(2+✓3)(2-✓5)+(2+✓3)(✓5-✓3)/(2+✓3)(✓5-✓3)(2-✓5)

(2✓5-5-2✓3+✓15)+2(4-2✓5+2✓3-✓15)+(2✓5-2✓3+✓15-✓3)/(2+✓3)(✓5-✓3)(2-✓5)

(4✓5-4✓5-4✓3+4✓3+2✓15-2✓15+3-3/(2+✓3)(✓5-✓3)(2-✓5)

0/(2+✓3)(✓5-✓3)(2-✓5)

0

L.H.S.=R.H.S.