Answer
The simplified form of the expression $\frac{1}{5-x}+\frac{x}{2x-10}$ is $\frac{\left( x-2 \right)}{2\left( x-5 \right)}$.
Work Step by Step
$\frac{1}{5-x}+\frac{x}{2x-10}$
Obtain the alternative form of the expression by multiplying the second term of rational expression $\frac{1}{5-x}+\frac{x}{2x-10}$ by 1 in the form$\frac{-1}{-1}$:
$\begin{align}
& \frac{1}{5-x}+\frac{x}{2x-10}=\frac{1}{5-x}+\frac{x}{2\left( x-5 \right)}.1 \\
& \frac{1}{5-x}+\frac{x}{2\left( x-5 \right)}.\frac{-1}{-1}
\end{align}$
Apply the Distributive property:
$\begin{align}
& \frac{1}{5-x}+\frac{x}{2x-10}=\frac{1}{5-x}+\frac{x\left( -1 \right)}{2\left( x-5 \right)\left( -1 \right)} \\
& =\frac{1}{5-x}+\frac{-\left( x \right)}{-2\left( x-5 \right)} \\
& =\frac{1}{5-x}-\frac{x}{2\left( 5-x \right)}
\end{align}$
$\begin{align}
& \frac{1}{5-x}=\frac{1\cdot 2}{2\left( 5-x \right)} \\
& =\frac{2}{2\left( 5-x \right)}
\end{align}$
And
$\frac{x}{2\left( x-5 \right)}=\frac{x}{2\left( x-5 \right)}$
Thus, expression becomes:
$\frac{1}{5-x}+\frac{x}{2x-10}=\frac{2}{2\left( 5-x \right)}-\frac{x}{2\left( x-5 \right)}$
Now, the denominators are same. So, subtract the numerators and keep the common denominator:
$\frac{1}{5-x}+\frac{x}{2x-10}=\frac{2-x}{2\left( 5-x \right)}$
Rearrange the terms and simplify further as follows:
$\begin{align}
& \frac{1}{5-x}+\frac{x}{2x-10}=\frac{2-x}{2\left( 5-x \right)} \\
& =\frac{x-2}{2\left( x-5 \right)}
\end{align}$