Answer
The required solution is $2$.
Work Step by Step
We have the given algebraic expression:
$\left( 2-\frac{6}{x+1} \right)\left( 1+\frac{3}{x-2} \right)$.
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:
$\begin{align}
& 2-\frac{6}{x+1}=2\times \frac{\left( x+1 \right)}{\left( x+1 \right)}-\frac{6}{\left( x+1 \right)} \\
& =\frac{2\left( x+1 \right)}{\left( x+1 \right)}-\frac{6}{\left( x+1 \right)} \\
& =\frac{2x+2-6}{\left( x+1 \right)} \\
& =\frac{2x-4}{\left( x+1 \right)}
\end{align}$.
And factorize the numerator of the obtained fraction to get a simplified form:
$2-\frac{6}{x+1}=\frac{2\left( x-2 \right)}{\left( x+1 \right)}$.
Solve the second bracket of the given rational expression:
$\begin{align}
& 1+\frac{3}{x-2}=1\times \frac{\left( x-2 \right)}{\left( x-2 \right)}+\frac{3}{\left( x-2 \right)} \\
& =\frac{x-2}{\left( x-2 \right)}+\frac{3}{\left( x-2 \right)} \\
& =\frac{x-2+3}{\left( x-2 \right)} \\
& =\frac{x+1}{\left( x-2 \right)}
\end{align}$.
Therefore, the given expression becomes
$\begin{align}
& \left( 2-\frac{6}{x+1} \right)\left( 1+\frac{3}{x-2} \right)=\frac{2\left( x-2 \right)}{\left( x+1 \right)}\times \frac{\left( x+1 \right)}{\left( x-2 \right)} \\
& =2
\end{align}$.
Hence, $\left( 2-\frac{6}{x+1} \right)\left( 1+\frac{3}{x-2} \right)=$ $2$.
