Answer
The simplified form of the complex rational expression $\frac{\frac{2}{x}-\frac{1}{{{x}^{2}}}}{\frac{x}{4}}$ is $\frac{4\left( 2x-1 \right)}{{{x}^{3}}}$.
Work Step by Step
$\frac{\frac{2}{x}-\frac{1}{{{x}^{2}}}}{\frac{x}{4}}$
The denominator of the complex rational expression has a single term as $\frac{x}{4}$.
But the numerator is not single, so perform the operation and get a single rational expression in the numerator:
The LCD of $\frac{2}{x}-\frac{1}{{{x}^{2}}}$ is ${{x}^{2}}$.
Now, solve the complex fraction by making all terms equivalent as its LCD:
$\begin{align}
& \frac{\frac{2}{x}-\frac{1}{{{x}^{2}}}}{\frac{x}{4}}=\frac{\frac{2}{x}\cdot \frac{{{x}^{2}}}{{{x}^{2}}}-\frac{1}{{{x}^{2}}}}{\frac{x}{4}} \\
& =\frac{\frac{2x}{{{x}^{2}}}-\frac{1}{{{x}^{2}}}}{\frac{x}{4}} \\
& =\frac{\frac{2x-1}{{{x}^{2}}}}{\frac{x}{4}}
\end{align}$
Divide the numerator by the denominator,
$\frac{\frac{2}{x}-\frac{1}{{{x}^{2}}}}{\frac{x}{4}}=\frac{2x-1}{{{x}^{2}}}\div \frac{x}{4}$
$\begin{align}
& \frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C} \\
& =\frac{AD}{BC}
\end{align}$
where $\frac{A}{B},\left( B\ne 0 \right)$, $\frac{C}{D},\left( D\ne 0 \right)$ are rational expressions with $\frac{C}{D}\ne 0$
The reciprocal of $\frac{x}{4}$ is $\frac{4}{x}$
Thus, the expression becomes:
$\frac{\frac{2}{x}-\frac{1}{{{x}^{2}}}}{\frac{x}{4}}=\frac{2x-1}{{{x}^{2}}}\cdot \frac{4}{x}$
Simplify the terms,
$\begin{align}
& \frac{\frac{2}{x}-\frac{1}{{{x}^{2}}}}{\frac{x}{4}}=\frac{\left( 2x-1 \right)\cdot 4}{\left( {{x}^{2}} \right)\left( x \right)} \\
& =\frac{4\left( 2x-1 \right)}{{{x}^{3}}}
\end{align}$
