Answer
$\dfrac{6m}{(m+2)(m-3)(2m+1)}$
Work Step by Step
Let's note:
$$E=\dfrac{5}{2m^2-5m-3}+\dfrac{3}{2m^2+5m+2}-\dfrac{1}{m^2-m-6}.$$
First factor all denominators:
$$\begin{align*}
E&=\dfrac{5}{(2m+1)(m-3)}+\dfrac{3}{(2m+1)(m+2)}-\dfrac{1}{(m+2)(m-3)}.
\end{align*}$$
The Least Common Denominator (LCD) is $(2m+1)(m-3)(m+2)$. Multiply the numerator and denominator of each fraction accordingly, then add numerators and place the sum over the Least Common Denominator:
$$\begin{align*}
E&=\dfrac{5(m+2)}{(2m+1)(m-3)(m+2)}+\dfrac{3(m-3)}{(2m+1)(m+2)(m-3)}-\dfrac{2m+1}{(m+2)(m-3)(2m+1)}\\
&=\dfrac{5m+10+3m-9-2m-1}{(m+2)(m-3)(2m+1)}\\
&=\dfrac{6m}{(m+2)(m-3)(2m+1)}.
\end{align*}$$
