If $x=2+\sqrt{3},xy=1$ , then $\frac{x}{\sqrt{2}+\sqrt{x}}+\frac{y}{\sqrt{2}-\sqrt{y}}=\sqrt{2}k$ , then $k=$

$\sqrt{x}=\sqrt{2+\sqrt{3}}$

$=\frac{\sqrt{2}}{\sqrt{2}}\times \sqrt{2+\sqrt{3}}$

$=\frac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}}$

$=\frac{\sqrt{3}+1}{\sqrt{2}}$

Also, given, $xy=1$

$\Rightarrow y=\frac{1}{x}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$

$\sqrt{y}=\sqrt{2-\sqrt{3}}$

$=\frac{\sqrt{2}}{\sqrt{2}}\times \sqrt{2-\sqrt{3}}$

$=\frac{1}{\sqrt{2}}\sqrt{4-\sqrt{3}}$

$=\frac{\sqrt{3}-1}{\sqrt{2}}$

To find

$\frac{x}{\sqrt{2}+\sqrt{x}}+\frac{y}{\sqrt{2}-\sqrt{y}}$

$=\frac{2+\sqrt{3}}{\sqrt{2}+\frac{1}{\sqrt{2}}\left(\sqrt{3}+1\right)}+\frac{2-\sqrt{3}}{\sqrt{2}-\frac{1}{\sqrt{2}}\left(\sqrt{3}-1\right)}$

$=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}$

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

$=\sqrt{2}$

$\Rightarrow \sqrt{2}k=\sqrt{2}$

$\therefore k=1$