Answer
The required solution is
$\frac{\left( 4x+5 \right)\left( x+4 \right)}{\left( x+2 \right)\left( x-1 \right)}$.
Work Step by Step
We have the given algebraic expression:
$\left( 4-\frac{3}{x+2} \right)\left( 1+\frac{5}{x-1} \right)$.
For an algebraic expression, a rational expression is an expression that can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator $q\ne 0$.
Now, solve the first bracket of the given rational expression:
$\begin{align}
& 4-\frac{3}{x+2}=4\times \frac{\left( x+2 \right)}{\left( x+2 \right)}-\frac{3}{\left( x+2 \right)} \\
& =\frac{4\left( x+2 \right)}{\left( x+2 \right)}-\frac{3}{\left( x+2 \right)} \\
& =\frac{4x+8-3}{\left( x+2 \right)} \\
& =\frac{4x+5}{\left( x+2 \right)}
\end{align}$.
Also, solve the second bracket of the given rational expression:
$\begin{align}
& 1+\frac{5}{x-1}=1\times \frac{\left( x-1 \right)}{\left( x-1 \right)}+\frac{5}{\left( x-1 \right)} \\
& =\frac{x-1}{\left( x-1 \right)}+\frac{5}{\left( x-1 \right)} \\
& =\frac{x-1+5}{\left( x-1 \right)} \\
& =\frac{x+4}{\left( x-1 \right)}
\end{align}$.
Therefore, the given expression becomes
$\begin{align}
& \left( 4-\frac{3}{x+2} \right)\left( 1+\frac{5}{x-1} \right)=\frac{\left( 4x+5 \right)}{\left( x+2 \right)}\times \frac{\left( x+4 \right)}{\left( x-1 \right)} \\
& =\frac{\left( 4x+5 \right)\left( x+4 \right)}{\left( x+2 \right)\left( x-1 \right)}
\end{align}$.
Hence, $\left( 4-\frac{3}{x+2} \right)\left( 1+\frac{5}{x-1} \right)=$ $\frac{\left( 4x+5 \right)\left( x+4 \right)}{\left( x+2 \right)\left( x-1 \right)}$.
