Question

proove that........​

Answer 1

Question :-

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

Solution :-

Take L. H. S.

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

Now rationalize every term

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

Now simplify this

$(\frac{2 - \sqrt{3} }{4 - 3}) + ( \frac{2 \sqrt{5} - 2 \sqrt{3} }{5 - 3} ) - ( \frac{ \sqrt{5} + 2} {5 - 4} ) \\ \\ \\ (2 - \sqrt{3} ) + ( \sqrt{5} - \sqrt{3} ) - ( \sqrt{5} + 2) \\ \\ \\ 2 - \sqrt{3} + \sqrt{5} - \sqrt{3} - \sqrt{5} - 2 = 0 \: \: \: \: \: \: \: \: \: rhs \\ \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: proved$

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

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

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

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

• rationalize the denominator

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

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

$- ( \dfrac{1}{ \sqrt{5} - 2} )\times(\dfrac{\sqrt 5 + 2}{\sqrt 5 + 2}) = 0$

___________________

$\sf (\dfrac{2 - \sqrt{3} }{4 - 3}) + ( \dfrac{2 \sqrt{5} - 2 \sqrt{3} }{5 - 3} ) - ( \dfrac{ \sqrt{5} + 2} {5 - 4} ) = 0$

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

$\sf \cancel 2 - \sqrt{3} + \sqrt{5} - \sqrt{3} - \sqrt{5} - \cancel 2 = 0$

$0 = 0$

$LHS = RHS$

......Hence Proved

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