Prove that: $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2-\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=\sqrt{2}$

(N/A) To simplify the expression,we first multiply the numerator and denominator of each term by $\sqrt{2}$ to eliminate the nested radicals.
Let the expression be $E = \frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2-\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}$.
Multiply the numerator and denominator of each term by $\sqrt{2}$:
$E = \frac{\sqrt{2}(2+\sqrt{3})}{\sqrt{2}(\sqrt{2}+\sqrt{2-\sqrt{3}})} + \frac{\sqrt{2}(2-\sqrt{3})}{\sqrt{2}(\sqrt{2}-\sqrt{2+\sqrt{3}})}$
$E = \frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4-2\sqrt{3}}} + \frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4+2\sqrt{3}}}$
Note that $4-2\sqrt{3} = (\sqrt{3}-1)^2$ and $4+2\sqrt{3} = (\sqrt{3}+1)^2$.
So,$\sqrt{4-2\sqrt{3}} = \sqrt{3}-1$ and $\sqrt{4+2\sqrt{3}} = \sqrt{3}+1$.
Substituting these values:
$E = \frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3}-1} + \frac{2\sqrt{2}-\sqrt{6}}{2-(\sqrt{3}+1)}$
$E = \frac{2\sqrt{2}+\sqrt{6}}{\sqrt{3}+1} + \frac{2\sqrt{2}-\sqrt{6}}{1-\sqrt{3}}$
$E = \frac{2\sqrt{2}+\sqrt{6}}{\sqrt{3}+1} - \frac{2\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}$
Taking the common denominator $(\sqrt{3}+1)(\sqrt{3}-1) = 3-1 = 2$:
$E = \frac{(2\sqrt{2}+\sqrt{6})(\sqrt{3}-1) - (2\sqrt{2}-\sqrt{6})(\sqrt{3}+1)}{2}$
$E = \frac{(2\sqrt{6}-2\sqrt{2}+3\sqrt{2}-\sqrt{6}) - (2\sqrt{6}+2\sqrt{2}-3\sqrt{2}-\sqrt{6})}{2}$
$E = \frac{(\sqrt{6}+\sqrt{2}) - (\sqrt{6}-\sqrt{2})}{2}$
$E = \frac{\sqrt{6}+\sqrt{2}-\sqrt{6}+\sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$.
Hence,the expression is proved to be $\sqrt{2}$.