#### Answer

$ \displaystyle \frac{1}{\sqrt{x+7}+\sqrt{x}}$

#### Work Step by Step

We lose the square roots in the numerator by applying the difference of squares formula:
$(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}$
$=a^{2}x-b^{2}y$
$\displaystyle \frac{\sqrt{x+7}-\sqrt{x}}{7}\color{red}{ \cdot\frac{\sqrt{x+7}+\sqrt{x}}{\sqrt{x+7}+\sqrt{x}} }\qquad$ (rationalize)
$=\displaystyle \frac{(\sqrt{x+7})^{2}-(\sqrt{x})^{2}}{7(\sqrt{x+7}+\sqrt{x})}$
$=\displaystyle \frac{x+7-x}{7(\sqrt{x+7}+\sqrt{x})}$
$=\displaystyle \frac{7}{7(\sqrt{x+7}+\sqrt{x})}$
$=\displaystyle \frac{1}{\sqrt{x+7}+\sqrt{x}}$