Answer
$\displaystyle \frac{1}{\sqrt{x+5}+\sqrt{x}}$
Work Step by Step
We lose the square roots in the numerator by applying the difference of squares formula:
$(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}$
$=a^{2}x-b^{2}y$
$\displaystyle \frac{\sqrt{x+5}-\sqrt{x}}{5}\color{red}{ \cdot\frac{\sqrt{x+5}+\sqrt{x}}{\sqrt{x+5}+\sqrt{x}} }\qquad$ (rationalize)
$=\displaystyle \frac{(\sqrt{x+5})^{2}-(\sqrt{x})^{2}}{5(\sqrt{x+5}+\sqrt{x})}$
$=\displaystyle \frac{x+5-x}{5(\sqrt{x+5}+\sqrt{x})}$
$=\displaystyle \frac{5}{5(\sqrt{x+5}+\sqrt{x})}$
$=\displaystyle \frac{1}{\sqrt{x+5}+\sqrt{x}}$
