Question

PLEASE ANSWER THE QUESTION IN THE ATTACHMENT

BRAINLIEST QUESTION

Answer 1

Please refer to your answer..
And if this helps mark brainiest

Answer 2

$Given : x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

On rationalizing, we get

$=> \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} * \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

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

$=> \frac{(3 + 2 + 2\sqrt{6})}{3 - 2}$

$=> (3 + 2 + 2\sqrt{6})$

-----------------------------------------------------------------------------------------------------------------

Now,

$=> \frac{x - y}{x - 3y}$

$=> \frac{3 + 2 + 2\sqrt{6} - 1 }{3 + 2 + 2\sqrt{6} - 3}$

$=> \frac{4 + 2\sqrt{6}}{2 + 2\sqrt{6} }$

$=> \frac{2(2 + \sqrt{6})}{2(1 + \sqrt{6}) }$

$=> \frac{2 + \sqrt{6}}{1 + \sqrt{6}}$

$=> \frac{2 + \sqrt{6} }{ 1 + \sqrt{6} } * \frac{1 - \sqrt{6} }{1 - \sqrt{6} }$

$=> \frac{(2 + \sqrt{6})(1 - \sqrt{6})}{1 - (\sqrt{6})^2 }$

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

$=> \frac{-4 - \sqrt{6}}{-5}$

$=> \frac{4 + \sqrt{6}}{5}$

Therefore, the final answer is:

$= > \boxed{\frac{4 + \sqrt{6}}{5}}$

Hope this helps!