Answer
$\dfrac{2(x+5)}{(x-1)(x+2)}$
Work Step by Step
Since the least common multiple (LCM) of the denominator is $(x-1)(x+2)$, then we have
$$
\frac{4}{x-1}-\frac{2}{x+2}=\frac{4(x+2)}{(x-1)(x+2)}-\frac{2(x-1)}{(x-1)(x+2)}
.$$
Now, since the denominators are the same, then we have
$$
\frac{4(x+2)}{(x-1)(x+2)}-\frac{2(x-1)}{(x-1)(x+2)}=\frac{4x+8-2x+2}{(x-1)(x+2)}\\
=\frac{2x+10}{(x-1)(x+2)}=\frac{2(x+5)}{(x-1)(x+2)}
.$$
