Solution.

Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0, \\ x \neq y\end{array}\right.$

$$
\begin{aligned}
& \frac{\frac{x+y}{\sqrt{x}-\sqrt{y}}-\frac{x-y}{\sqrt{x}+\sqrt{y}}}{\frac{\sqrt{x}-\sqrt{y}}{x+y}+\frac{\sqrt{x}+\sqrt{y}}{x-y}} \cdot \frac{y-\sqrt{x y}+x}{2 \sqrt{x y}}= \\
& =\frac{\frac{(x+y)(\sqrt{x}+\sqrt{y})-(x-y)(\sqrt{x}-\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}}{\frac{(\sqrt{x}-\sqrt{y})(x-y)+(\sqrt{x}+\sqrt{y})(x+y)}{(x+y)(x-y)}} \cdot \frac{y-\sqrt{x y}+x}{2 \sqrt{x y}}= \\
& x \sqrt{x}+x \sqrt{y}+y \sqrt{x}+y \sqrt{y}-x \sqrt{x}+x \sqrt{y}+y \sqrt{x}-y \sqrt{y} \\
& =\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\frac{x \sqrt{x}-x \sqrt{y}-y \sqrt{x}+y \sqrt{y}+x \sqrt{x}+x \sqrt{y}+y \sqrt{x}-y \sqrt{y}}{(x+y)(x-y)}} \cdot \frac{y-\sqrt{x y}+x}{2 \sqrt{x y}}= \\
& =\frac{2 \sqrt{x y}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})} \cdot \frac{(x+y)(x-y)}{2(x \sqrt{x}+y \sqrt{y})} \cdot \frac{y-\sqrt{x y}+x}{2 \sqrt{x y}}=
\end{aligned}
$$

$$
=\frac{2 \sqrt{x y}}{\sqrt{x}-\sqrt{y}} \cdot \frac{(x+y)(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{2(\sqrt{x}+\sqrt{y})(x-\sqrt{x y}+y)} \cdot \frac{y-\sqrt{x y}+x}{2 \sqrt{x y}}=\frac{x+y}{2} .
$$

Answer: $\frac{x+y}{2}$.
2.171. $\frac{\frac{x+y}{\sqrt{x}-\sqrt{y}}-\frac{x-y}{\sqrt{x}+\sqrt{y}}}{\frac{\sqrt{x}-\sqrt{y}}{x+y}+\frac{\sqrt{x}+\sqrt{y}}{x-y}} \cdot \frac{y-\sqrt{x y}+x}{2 \sqrt{x y}}$.