Answer
The required solution is:
$\frac{{{x}^{2}}+5x+8}{\left( x+2 \right)\left( x+1 \right)}$.
Work Step by Step
We have the given algebraic expression:
$\left( \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3} \right)-\frac{2}{x+2}$.
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 bracket of the above rational expression:
$\begin{align}
& \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3}=\frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+\left( 5-1 \right)x-5}{2{{x}^{2}}+\left( 3-2 \right)x-3} \\
& =\frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+5x-1x-5}{2{{x}^{2}}+3x-2x-3} \\
& =\frac{2x+3}{x+1}\cdot \frac{x\left( x+5 \right)-1\left( x+5 \right)}{x\left( 2x+3 \right)-1\left( 2x+3 \right)} \\
& =\frac{2x+3}{x+1}\cdot \frac{\left( x+5 \right)\left( x-1 \right)}{\left( 2x+3 \right)\left( x-1 \right)}
\end{align}$.
Solving further,
$\begin{align}
& \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3}=\frac{\left( 2x+3 \right)}{\left( x+1 \right)}\cdot \frac{\left( x+5 \right)\left( x-1 \right)}{\left( 2x+3 \right)\left( x-1 \right)} \\
& =\frac{x+5}{x+1}
\end{align}$.
So, the given expression becomes
$\left( \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3} \right)-\frac{2}{x+2}=\frac{x+5}{x+1}-\frac{2}{x+2}$.
And simplify the above expression by taking the least common denominator:
$\begin{align}
& \frac{x+5}{x+1}-\frac{2}{x+2}=\frac{x+5}{\left( x+1 \right)}\times \frac{\left( x+2 \right)}{\left( x+2 \right)}-\frac{2}{\left( x+2 \right)}\times \frac{\left( x+1 \right)}{\left( x+1 \right)} \\
& =\frac{\left( x+5 \right)\left( x+2 \right)}{\left( x+1 \right)\left( x+2 \right)}-\frac{2\left( x+1 \right)}{\left( x+2 \right)\left( x+1 \right)} \\
& =\frac{{{x}^{2}}+5x+2x+10-2x-2}{\left( x+2 \right)\left( x+1 \right)} \\
& =\frac{{{x}^{2}}+5x+8}{\left( x+2 \right)\left( x+1 \right)}
\end{align}$
Therefore, the given expression gets reduced to
$\left( \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3} \right)-\frac{2}{x+2}=\frac{{{x}^{2}}+5x+8}{\left( x+2 \right)\left( x+1 \right)}$
Hence, $\left( \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3} \right)-\frac{2}{x+2}=$ $\frac{{{x}^{2}}+5x+8}{\left( x+2 \right)\left( x+1 \right)}$.
