Answer
The simplified form of the rational expression $\frac{2x}{{{x}^{2}}-3x}\div \left( x-3 \right)$ is$\frac{2}{{{\left( x-3 \right)}^{2}}}$.
Work Step by Step
$\frac{2x}{{{x}^{2}}-3x}\div \left( x-3 \right)$
$\frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C}$
The reciprocal of $\frac{\left( x-3 \right)}{1}$ is $\frac{1}{\left( x-3 \right)}$
So, multiply the reciprocal of the divisor,
$\begin{align}
& \frac{2x}{{{x}^{2}}-3x}\div \left( x-3 \right)=\frac{2x}{{{x}^{2}}-3x}\cdot \frac{1}{\left( x-3 \right)} \\
& =\frac{\left( 2x \right)1}{\left( {{x}^{2}}-3x \right)\left( x-3 \right)}
\end{align}$
$\left( {{x}^{2}}-3x \right)\left( x-3 \right)=x\left( x-3 \right)\left( x-3 \right)$
So, the rational expression becomes,
$\frac{2x}{{{x}^{2}}-3x}\div \left( x-3 \right)=\frac{\left( 2x \right)}{x\left( x-3 \right)\left( x-3 \right)}$
Regroup and remove the factor equal to 1,
$\begin{align}
& \frac{2x}{{{x}^{2}}-3x}\div \left( x-3 \right)=\frac{2x}{x\left( x-3 \right)\left( x-3 \right)} \\
& =\frac{x}{x}\cdot \frac{2}{\left( x-3 \right)\left( x-3 \right)} \\
& =1\cdot \frac{2}{{{\left( x-3 \right)}^{2}}} \\
& =\frac{2}{{{\left( x-3 \right)}^{2}}}
\end{align}$
