23. Упростите выражение $$(\frac{1}{\sqrt{y}}-\frac{2}{\sqrt{x}+\sqrt{y}}):(\sqrt{x}-\frac{x+y}{\sqrt{x}+\sqrt{y}})$$

Упростим выражение.

$$(\frac{1}{\sqrt{y}}-\frac{2}{\sqrt{x}+\sqrt{y}}):(\sqrt{x}-\frac{x+y}{\sqrt{x}+\sqrt{y}}) = (\frac{\sqrt{x}+\sqrt{y}-2\sqrt{y}}{\sqrt{y}(\sqrt{x}+\sqrt{y})}):(\frac{\sqrt{x}(\sqrt{x}+\sqrt{y})-(x+y)}{\sqrt{x}+\sqrt{y}}) = (\frac{\sqrt{x}-\sqrt{y}}{\sqrt{y}(\sqrt{x}+\sqrt{y})}):(\frac{x+\sqrt{xy}-x-y}{\sqrt{x}+\sqrt{y}}) = (\frac{\sqrt{x}-\sqrt{y}}{\sqrt{y}(\sqrt{x}+\sqrt{y})}):(\frac{\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}) = (\frac{\sqrt{x}-\sqrt{y}}{\sqrt{y}(\sqrt{x}+\sqrt{y})})\cdot(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y}(\sqrt{x}-\sqrt{y})}) = \frac{1}{\sqrt{y}} \cdot \frac{1}{\sqrt{y}} = \frac{1}{y}$$.

Ответ: $$\frac{1}{y}$$