Question

prove that 1/2+ root 3 + 2/root t - root 3 + 1/2- root 5​

Answer 1

Topic :-

Surds

Correct Question :-

Prove that :-

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

SOLUTION:-

Take L.C.M to the denominator

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

$\dfrac{2 \sqrt{5} - 5 - 2 \sqrt{3} + \sqrt{15} + 2(4 - 2 \sqrt{5} + 2 \sqrt{3} - \sqrt{15} ) + 2 \sqrt{5} - 2 \sqrt{3} + \sqrt{15} - 3 }{(2 + \sqrt{3} )( \sqrt{5} - \sqrt{3})(2 - \sqrt{5} ) }$

Keeping the like terms together in the numerator

$\dfrac{2 \sqrt{5} + 2 \sqrt{5} - 4 \sqrt{5} - 2 \sqrt{3} - 2 \sqrt{3} + 4 \sqrt{3} + \sqrt{15} + \sqrt{15} - 2 \sqrt{15} - 5 - 3 + 8 }{(2 + \sqrt{3})( \sqrt{5 } - \sqrt{3})(2 - \sqrt{5}) }$

$\dfrac{4 \sqrt{5} - 4 \sqrt{5} - 4 \sqrt{3} + 4 \sqrt{3} + 2 \sqrt{15} - 2 \sqrt{15} - 8 + 8 }{(2 + \sqrt{3})( \sqrt{5} - \sqrt{3} )(2 - \sqrt{5} ) }$

All terms get cancelled in the numerator

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

$= 0$

Hence proved!

So,

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

Note :-

If you are confusion to seè the answer .Kindly refer the attachment