Answer
$\dfrac{(4x+5)(x+4)}{(x+2)(x-1)}$
Work Step by Step
Let's note:
$$E=\left(4-\dfrac{3}{x+2}\right)\left(1+\dfrac{5}{x-1}\right)$$
First perform the operations inside the parenthesis:
$$\begin{align*}
E&=\left(\dfrac{4(x+2)-3}{x+2}\right)\left(\dfrac{x-1+5}{x-1}\right)\\
&=\dfrac{4x+8-3}{x+2}\cdot\dfrac{x+4}{x-1}\\
&=\dfrac{4x+5}{x+2}\cdot\dfrac{x+4}{x-1}.
\end{align*}$$
Perform multiplication:
$$\begin{align*}
E&=\dfrac{(4x+5)(x+4)}{(x+2)(x-1)}.
\end{align*}$$
