Answer
The simplified form of the rational expression $\frac{2-x}{5{{x}^{2}}}\div \frac{{{x}^{2}}-4}{3x}$ is$-\frac{3}{5x\cdot \left( x+2 \right)}$.
Work Step by Step
$\frac{2-x}{5{{x}^{2}}}\div \frac{{{x}^{2}}-4}{3x}$
$\frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C}$
The reciprocal of $\frac{{{x}^{2}}-4}{3x}$ is$\frac{3x}{{{x}^{2}}-4}$.
So, multiply the reciprocal of the divisor,
$\begin{align}
& \frac{2-x}{5{{x}^{2}}}\div \frac{{{x}^{2}}-4}{3x}=\frac{2-x}{5{{x}^{2}}}\cdot \frac{3x}{{{x}^{2}}-4} \\
& =\frac{\left( 2-x \right)\left( 3x \right)}{5{{x}^{2}}\left( {{x}^{2}}-4 \right)}
\end{align}$
Factor the terms,
$\begin{align}
& \left( 2-x \right)\left( 3x \right)=-\left( x-2 \right)\left( 3x \right) \\
& 5{{x}^{2}}\left( {{x}^{2}}-4 \right)=5{{x}^{2}}\left( {{x}^{2}}-{{2}^{2}} \right) \\
& =5{{x}^{2}}\left( x+2 \right)\left( x-2 \right)
\end{align}$
So, the rational expression becomes,
$\frac{2-x}{5{{x}^{2}}}\div \frac{{{x}^{2}}-4}{3x}=\frac{-\left( x-2 \right)\left( 3x \right)}{5{{x}^{2}}\left( x+2 \right)\left( x-2 \right)}$
Regroup and remove the factor equal to 1,
$\begin{align}
& \frac{2-x}{5{{x}^{2}}}\div \frac{{{x}^{2}}-4}{3x}=\frac{-\left( x-2 \right)\left( 3x \right)}{\left( x-2 \right)\left( x+2 \right)5{{x}^{2}}} \\
& =\frac{-\left( x-2 \right)}{\left( x-2 \right)}\cdot \frac{\left( 3x \right)}{\left( x+2 \right)5{{x}^{2}}} \\
& =-1\cdot \frac{3x}{\left( x+2 \right)5{{x}^{2}}} \\
& =-\frac{3}{5x\cdot \left( x+2 \right)}
\end{align}$