Answer
The simplified form of the complex rational expression $\frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}$ is $-\frac{1}{x}$.
Work Step by Step
$\frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}$
The denominator of the complex rational expression is single.
But, the numerator is not single, so perform the operation and get a single rational expression in the numerator.
The LCD of $\frac{1}{2}-\frac{1}{x}$ is $2x$.
Solve the expression by making all fractions equivalent to their LCD:
$\begin{align}
& \frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{\frac{1}{2}\cdot \frac{x}{x}-\frac{1}{x}\cdot \frac{2}{2}}{\frac{2-x}{2}} \\
& =\frac{\frac{x}{2x}-\frac{2}{2x}}{\frac{2-x}{2}} \\
& =\frac{\frac{x-2}{2x}}{\frac{2-x}{2}}
\end{align}$
Divide the numerator by the denominator,
$\frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{x-2}{2x}\div \frac{2-x}{2}$
$\begin{align}
& \frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C} \\
& =\frac{AD}{BC}
\end{align}$
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 $\frac{2-x}{2}$ is $\frac{2}{2-x}$
So, multiply the reciprocal of the divisor,
$\frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{x-2}{2x}\cdot \frac{2}{2-x}$
Simplify the terms,
$\begin{align}
& \frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{\left( x-2 \right)\cdot 2}{\left( 2x \right)\left( 2-x \right)} \\
& =\frac{2\left( x-2 \right)}{2x\left( 2-x \right)} \\
& =\frac{x-2}{x\left( 2-x \right)}
\end{align}$
Take out ‘–’ from the numerator,
$\begin{align}
& \frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{-\left( -x+2 \right)}{x\left( 2-x \right)} \\
& =\frac{-\left( 2-x \right)}{x\left( 2-x \right)} \\
& =-\frac{1}{x}
\end{align}$
