Answer
$\displaystyle \frac{a-b}{a+2\sqrt{ab}+b}$
Work Step by Step
We lose the square roots in the numerator by applying the difference of squares formula:
$(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}$
$=a^{2}x-b^{2}y$
$\displaystyle \frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\color{red}{ \cdot\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}} }\qquad$ (rationalize)
$=\displaystyle \frac{(\sqrt{a})^{2}-(\sqrt{b})^{2}}{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})}$
... the denominator is a square of a sum, $(A+B)^{2}=A^{2}+2AB+B^{2}$
$=\displaystyle \frac{a-b}{(\sqrt{a})^{2}+2\sqrt{ab}+(\sqrt{b})^{2}}$
$=\displaystyle \frac{a-b}{a+2\sqrt{ab}+b}$
