Question

plz plz ans marks r 10+ i will mark brainlest​

Answer 1

Answer:

99

Step-by-step explanation:

First, notice that y is just the reciprocal of x, so

   xy = 1.

Next, notice that

 x² + y² + xy = ( x² + y² - 2xy ) + 3xy = ( y - x )² + 3

so we just need to get y-x.

Now

$y - x = \frac{\textstyle\sqrt3+\sqrt2}{\textstyle\sqrt3-\sqrt2} - \frac{\textstyle\sqrt3-\sqrt2}{\textstyle\sqrt3+\sqrt2}\\\\=\frac{\textstyle(\sqrt3+\sqrt2)^2-(\sqrt3-\sqrt2)^2}{\textstyle(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\\\\=\frac{\textstyle(3+2+2\sqrt6)-(3+2-2\sqrt6)}{\textstyle3-2}\\\\=4\sqrt6$

So, finally, we get

 x² + y² + xy = ( y - x )² + 3 = (4√6)² + 3 = 16×6 + 3 = 96 + 3 = 99

Hope this helps you!

Answer 2

Answer: 99

Step-by-step explanation:

$x =\frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} +\sqrt{2} } \\x =\frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} +\sqrt{2} }*\frac{\sqrt{3} -\sqrt{2} }{\sqrt{3} -\sqrt{2} }$        (multiplying and dividing by $\sqrt{3} -\sqrt{2}$)

$x =\frac{(\sqrt{3} - \sqrt{2})^2 }{(\sqrt{3})^2 -(\sqrt{2})^2 }$              ( ∵ $(a+b)(a-b)= a^2-b^2$ )

$x =\frac{(\sqrt{3} - \sqrt{2})^2 }{(3 -2) }$

$x ={(\sqrt{3} - \sqrt{2})^2 }$

$x = 3+2-2*\sqrt{3} *\sqrt{2}$          (∵ $(a-b)^2 = a^2 + b^2 -2ab$ )

$x = 5-2\sqrt{6}$  

similarly, doing same for y, we get

$y = (\sqrt{3} +\sqrt{2} )^2$

$y= 5 + 2\sqrt{6}$

now, $x^2+y^2+xy = (x+y)^2 -xy$   (∵  $(x+y)^2 = x^2 + y^2 +2xy$ ) .... (1)

$x+y = 5 -2\sqrt{6} + 5 +2\sqrt{6} = 5+5 = 10$   and

$xy= (5+2\sqrt{6} )(5-2\sqrt{6} ) = (5)^2 -(2\sqrt{6} )^2 = 25-4*6 =25-24=1$

so putting $x+y=10$ and $xy=1$ in equation (1)

$x^2+y^2+xy = (x+y)^2 -xy$

                     $= (10)^2-1\\= 100-1\\=99$

so $x^2+y^2+xy = 99$