Question

Prove that 1/(2+ sqrt 3) + 2/(sqrt 5 - sqrt 3) +1/ (2- sqrt 5) =0

Answer 1

1/(2+sqr3)+2/(sqr5-sq3)+1/(2-sqr5)
taking LCM 
= (sqr5-sqr3)(2-sqr5)+2(2+sqr3)(2-sq5)+(2+sqr3)(sqr5-sqr3)/(2+sqr3)(sqr5-        sqr3)(2-sqr5)
= 2sqr5-5-2sqr3+sqr15+8-4sqr5+4sqr3-2sqr15+2sqr5-3+sqr15-2sqr3/......
= -5+8-3/.....
= 0/.....
0 by Any thing is 0
Hence Proved 

Answer 2

We need to recall the following definition of conjugate.

Conjugate is a change in the sign in the middle of two terms.

($+$ to $-$ ) or ($-$ to $+$)

This problem is about the conjugation of a term.

Given:

Prove that: $\frac{1}{2+\sqrt{3} }+\frac{2}{\sqrt{5} -\sqrt{3} }+\frac{1}{2-\sqrt{5} } =0$

Let's consider

$\frac{1}{2+\sqrt{3} }+\frac{2}{\sqrt{5} -\sqrt{3} }+\frac{1}{2-\sqrt{5} }$

Multiply the numerator and denominator of each term by the conjugate of the denominator.

$=\frac{1(2-\sqrt{3} )}{(2+\sqrt{3})(2-\sqrt{3} ) }+\frac{2(\sqrt{5} +\sqrt{3})}{(\sqrt{5} -\sqrt{3})(\sqrt{5} +\sqrt{3}) }+\frac{1(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5}) }$

$=\frac{2-\sqrt{3} }{4-3 }+\frac{2(\sqrt{5} +\sqrt{3})}{5-3 }+\frac{2+\sqrt{5}}{4-5 }$

$=\frac{2-\sqrt{3} }{1 }+\frac{2(\sqrt{5} +\sqrt{3})}{2 }+\frac{2+\sqrt{5}}{-1 }$

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

$=2-2-\sqrt{3}+\sqrt{3}+\sqrt{5}- \sqrt{5}$

$=0$

Hence, proved.