Answer
$\frac{4-x}{\sqrt{x+4}(x+4)}$
Work Step by Step
First, re-write to make it in radical form:
$\frac{{\sqrt{x+4}}-\frac{2x}{\sqrt{x+4}}}{x+4}$
Then, make both denominators of the top fraction the same:
$\frac{\frac{\sqrt{x+4}\cdot\sqrt{x+4}}{\sqrt{x+4}}-\frac{2x}{\sqrt{x+4}}}{x+4}$
Because of the law of radicals $\sqrt[n] a \cdot \sqrt[n] b=\sqrt[n] {ab}$:
$\frac{\frac{\sqrt{(x+4)^2}}{\sqrt{x+4}}-\frac{2x}{\sqrt{x+4}}}{x+4}$
Simplify further:
$\frac{\frac{x+4-2x}{\sqrt{x+4}}}{x+4}=\frac{4-x}{\sqrt{x+4}}\cdot\frac{1}{x+4}=\frac{4-x}{\sqrt{x+4}(x+4)}$
