Question

please answer to this question
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Answer 1

Given

$x = \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } \: \: \: \: \: y = \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1}$

And

$x ^{2} + y^{2} + xy$

Put in the values of x and y

$= ( \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } ) ^{2} + ( \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 } )^{2} + ( \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } \times \frac{ \sqrt{2 } - 1 }{ \sqrt{2 } + 1 } )$

By solving we get

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

Root to changes to 2 because root 2 x root 2 is 2.

Then it is

$3 + \frac{1}{3} + 1$
$= \frac{7}{3}$
Hope it helps