#### Answer

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

#### Work Step by Step

$\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}$
Multiply the numerator and the denominator of this expression by the conjugate of the numerator and simplify if possible:
$\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{(\sqrt{x})^{2}-(\sqrt{y})^{2}}{(\sqrt{x}-\sqrt{y})^{2}}=...$
$...=\dfrac{x-y}{(\sqrt{x})^{2}-2\sqrt{xy}+(\sqrt{y})^{2}}=\dfrac{x-y}{x-2\sqrt{xy}+y}$