Answer
$\frac{1}{(x+y)(\sqrt x- \sqrt y)}$.
Work Step by Step
The given expression is
$=\frac{\sqrt x+ \sqrt y}{x^2-y^2}$
Use the special formula $A^2-B^2=(A+B)(A-B)$.
$=\frac{\sqrt x+ \sqrt y}{(x+y)(x-y)}$
The conjugate of the numerator is $\sqrt x- \sqrt y$.
Multiply the numerator and the denominator by $\sqrt x- \sqrt y$.
$=\frac{\sqrt x+ \sqrt y}{(x+y)(x-y)}\cdot \frac{\sqrt x- \sqrt y}{\sqrt x- \sqrt y}$
Use the special formula $(A+B)(A-B)=A^2-B^2$.
$=\frac{(\sqrt x)^2- (\sqrt y)^2}{(x+y)(x-y)(\sqrt x- \sqrt y)}$
Simplify.
$=\frac{x -y}{(x+y)(x-y)(\sqrt x- \sqrt y)}$
Cancel common terms.
$=\frac{1}{(x+y)(\sqrt x- \sqrt y)}$.
