Question

REFER TO THE ATTACHMENT​

Answer 1

Answer:

99

Step-by-step explanation:

use the concept of surds

Answer 2

Answer:

Step-by-step explanation:

We have,

$\tt{x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\,\,\,\,\,and\,\,\,\,\,y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

Now, add the given expression,

$\tt{x+y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

$\tt{\implies\,x+y=\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2+\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}}$

$\tt{\implies\,x+y=\dfrac{2\left\{\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2\right\}}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}}$

$\tt{\implies\,x+y=\dfrac{2\left\{3+2\right\}}{3-2}}$

$\tt{\implies\,x+y=\dfrac{2(5)}{1}=10}$

And,

$\tt{xy=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\cdot\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=1}$

Now,

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

$\sf{=x^2+y^2+2xy-xy}$

$\sf{=(x+y)^2-xy}$

$\sf{=(10)^2-1=99}$