Answer
$\frac{-3}{\left( x+4 \right)}$.
Work Step by Step
$\frac{1}{4-x}+\frac{8}{{{x}^{2}}-16}-\frac{2}{x+4}$,
The expression can be written as,
$\frac{1}{4-x}+\frac{8}{{{x}^{2}}-16}-\frac{2}{x+4}=\frac{-1}{x-4}+\frac{8}{{{x}^{2}}-{{4}^{2}}}-\frac{2}{x+4}$
Apply the property ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ in the second expression,
$\frac{1}{4-x}+\frac{8}{{{x}^{2}}-16}-\frac{2}{x+4}=\frac{-1}{x-4}+\frac{8}{\left( x-4 \right)\left( x+4 \right)}-\frac{2}{x+4}$
LCD (Least Common Denominator) of the denominator is
$\text{LCD}=\left( x-4 \right)\left( x+4 \right)$
$\begin{align}
& \frac{1}{4-x}+\frac{8}{{{x}^{2}}-16}-\frac{2}{x+4}=\frac{-1}{x-4}\cdot \frac{x+4}{x+4}+\frac{8}{{{x}^{2}}-16}-\frac{2}{x+4}\cdot \frac{x-4}{x-4} \\
& =\frac{-\left( x+4 \right)}{\left( x-4 \right)\left( x+4 \right)}+\frac{8}{\left( x-4 \right)\left( x+4 \right)}-\frac{2\left( x-4 \right)}{\left( x-4 \right)\left( x+4 \right)} \\
& =\frac{-\left( x+4 \right)+8-2\left( x-4 \right)}{\left( x-4 \right)\left( x+4 \right)}
\end{align}$
Apply the distributive property and simplify the expression,
$\begin{align}
& \frac{1}{4-x}+\frac{8}{{{x}^{2}}-16}-\frac{2}{x+4}=\frac{-x-4+8-2x+8}{\left( x-4 \right)\left( x+4 \right)} \\
& =\frac{-3x+12}{\left( x-4 \right)\left( x+4 \right)} \\
& =\frac{-3\left( x-4 \right)}{\left( x-4 \right)\left( x+4 \right)} \\
& =\frac{-3}{\left( x+4 \right)}
\end{align}$
Therefore, the simplified form of the expression is $\frac{-3}{\left( x+4 \right)}$.
