Answer
The rationalized form of the expression $\frac{\sqrt{x+7}-\sqrt{x}}{7}$ is $\frac{1}{\left( \sqrt{x+7}+\sqrt{x} \right)}$.
Work Step by Step
Consider the provided expression, $\frac{\sqrt{x+7}-\sqrt{x}}{7}$
Multiply the numerator and denominator by $\sqrt{x+7}+\sqrt{x}$ so that the radicand in the numerator is a
perfect square.
Therefore,
$\frac{\sqrt{x+7}-\sqrt{x}}{7}=\frac{\sqrt{x+7}-\sqrt{x}}{7}\cdot \frac{\sqrt{x+7}+\sqrt{x}}{\sqrt{x+7}+\sqrt{x}}$
Apply the difference of squares property in the numerator,
$\left( \sqrt{x+7}+\sqrt{x} \right)\left( \sqrt{x+7}-\sqrt{x} \right)={{\left( \sqrt{x+7} \right)}^{2}}-{{\left( \sqrt{x} \right)}^{2}}$
Apply the power of a power property ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ in the expression ${{\left( \sqrt{x+7} \right)}^{2}},{{\left( \sqrt{x} \right)}^{2}}$.
$\begin{align}
& {{\left( \sqrt{x+7} \right)}^{2}}={{\left( x+7 \right)}^{2\cdot \frac{1}{2}}} \\
& ={{\left( x+7 \right)}^{1}} \\
& =x+7
\end{align}$
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:
$\begin{align}
& \frac{\sqrt{x+7}-\sqrt{x}}{7}=\frac{\sqrt{x+7}-\sqrt{x}}{7}\cdot \frac{\sqrt{x+7}+\sqrt{x}}{\sqrt{x+7}+\sqrt{x}} \\
& =\frac{{{\left( \sqrt{x+7} \right)}^{2}}-{{\left( \sqrt{x} \right)}^{2}}}{7\left( \sqrt{x+7}+\sqrt{x} \right)} \\
& =\frac{x+7-x}{7\left( \sqrt{x+7}+\sqrt{x} \right)} \\
& =\frac{7}{7\left( \sqrt{x+7}+\sqrt{x} \right)}
\end{align}$
Further simplify,
$\begin{align}
& \frac{\sqrt{x+7}-\sqrt{x}}{7}=\frac{7}{7\left( \sqrt{x+7}+\sqrt{x} \right)} \\
& =\frac{1}{\left( \sqrt{x+7}+\sqrt{x} \right)}
\end{align}$
Therefore, the rationalized form of the expression $\frac{\sqrt{x+7}-\sqrt{x}}{7}$ is $\frac{1}{\left( \sqrt{x+7}+\sqrt{x} \right)}$.
