#### Answer

$\dfrac{x-y}{x+2\sqrt{xy}+y}$

#### Work Step by Step

Multiplying by the conjugate of the numerator, the rationalized-numerator form of the given expression, $
\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}
\\\\=
\dfrac{(\sqrt{x})^2-(\sqrt{y})^2}{\sqrt{x}(\sqrt{x})+\sqrt{x}(\sqrt{y})+\sqrt{y}(\sqrt{x})+\sqrt{y}(\sqrt{y})}
\\\\=
\dfrac{x-y}{\sqrt{x(x)}+\sqrt{x(y)}+\sqrt{y(x)}+\sqrt{y(y)}}
\\\\=
\dfrac{x-y}{\sqrt{(x)^2}+\sqrt{xy}+\sqrt{xy}+\sqrt{(y)^2}}
\\\\=
\dfrac{x-y}{x+2\sqrt{xy}+y}
.\end{array}