NMTC 2015

Algebra Level 2

a a + b b ( a + b ) ( a − b ) + 2 b a + b − a b a − b \large{\dfrac { a\sqrt { a } +b\sqrt { b } }{ (\sqrt { a } +\sqrt { b } ) (a-b) } + \dfrac { 2\sqrt { b } }{ \sqrt { a } +\sqrt { b } } -\dfrac { \sqrt { ab } }{ a-b }}

If a = 2015 a=2015 and b = 2016 b=2016 , find the value of the expression above.

1 0 2016 \sqrt { 2016 } ( 2015 ) 2 { (2015) }^{ 2 }

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Bala Vidyadharan by Bala vidyadharan , 20, India

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2 solutions

Aaaaa Bbbbb
Aug 24, 2015

a a + b b + 2 b ( a − b ) − a b ( a + b ) a\sqrt{a}+b\sqrt{b}+2\sqrt{b}(a-b)-\sqrt{ab}(\sqrt{a}+\sqrt{b}) a a + b b + 2 a b − 2 b b − a b − b a a\sqrt{a}+b\sqrt{b}+2a\sqrt{b}-2b\sqrt{b}-a\sqrt{b}-b\sqrt{a} a a − b b + a b − b a a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a} a ( a − b ) + b ( a − b ) = ( a − b ) ( a + b ) \sqrt{a}(a-b)+\sqrt{b}(a-b)=(a-b)(\sqrt{a}+\sqrt{b}) ⇒ r e s = 1 \Rightarrow res=\boxed{1}

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Shreeprada Hegde
Aug 29, 2015

To solve this question take L.C.M of the denominator.that is( √a+√b)(a-b).And solve the equation that you get after taking L.C.M.You get neumartor and denominator same.So answer is 1

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