Answer
The rationalized form of the expression $\frac{\sqrt{x}-\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}$ is $\frac{1}{\left( x+y \right)\left( \sqrt{x}+\sqrt{y} \right)},x\ne y$.
Work Step by Step
Consider the provided expression, $\frac{\sqrt{x}-\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}$
Multiply the numerator and denominator by $\sqrt{x}+\sqrt{y}$ so that the radicand in the numerator is a
perfect square.
$\frac{\sqrt{x}-\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}=\frac{\sqrt{x}-\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}\cdot \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
Apply the difference of squares property in the numerator and denominator,
$\left( \sqrt{x}+\sqrt{y} \right)\left( \sqrt{x}-\sqrt{y} \right)={{\left( \sqrt{x} \right)}^{2}}-{{\left( \sqrt{y} \right)}^{2}}$
And,
${{x}^{2}}-{{y}^{2}}=\left( x-y \right)\left( x+y \right)$
Apply the power of a power property ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ in the expression ${{\left( \sqrt{y} \right)}^{2}},{{\left( \sqrt{x} \right)}^{2}}$.
Therefore,
$\begin{align}
& {{\left( \sqrt{y} \right)}^{2}}={{\left( y \right)}^{2\cdot \frac{1}{2}}} \\
& ={{\left( y \right)}^{1}} \\
& =y
\end{align}$
And,
$\begin{align}
& {{\left( \sqrt{x} \right)}^{2}}={{\left( x \right)}^{2\cdot \frac{1}{2}}} \\
& ={{\left( x \right)}^{1}} \\
& =x
\end{align}$
Rewrite the original expression as:
$\begin{align}
& \frac{\sqrt{x}-\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}=\frac{\sqrt{x}-\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}\cdot \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}} \\
& =\frac{{{\left( \sqrt{x} \right)}^{2}}-{{\left( \sqrt{y} \right)}^{2}}}{\left( {{x}^{2}}-{{y}^{2}} \right)\left( \sqrt{x}+\sqrt{y} \right)} \\
& =\frac{x-y}{\left( {{x}^{2}}-{{y}^{2}} \right)\left( \sqrt{x}+\sqrt{y} \right)} \\
& =\frac{x-y}{\left( {{x}^{2}}-{{y}^{2}} \right)\left( \sqrt{x}+\sqrt{y} \right)}
\end{align}$
Further simplify,
$\begin{align}
& \frac{\sqrt{x}-\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}=\frac{\left( x-y \right)}{\left( x-y \right)\left( x+y \right)\left( \sqrt{x}+\sqrt{y} \right)} \\
& =\frac{1}{\left( x+y \right)\left( \sqrt{x}+\sqrt{y} \right)}
\end{align}$
Therefore, the rationalized form of the expression $\frac{\sqrt{x}-\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}$ is $\frac{1}{\left( x+y \right)\left( \sqrt{x}+\sqrt{y} \right)}$.
