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