Question

If x=√3+√2/√3-√2,y=√3-√2/√3+√2 then find x+y.

Answer 1

HELLO DEAR,

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = > x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = > x = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2 } = {5 + 2 \sqrt{6} }$

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$y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = > y = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} = 5 - 2 \sqrt{6}$

now ,

x+y = (5+2√6) + ( 5 - 2√6)

=> 5+2√6 + 5 - 2√6

=> 5+5 +2√6 -2√6

=> 5+5 = 10

I HOPE ITS HELP YOU DEAR,
THANKS