Answer
The required value is $\frac{\left( x+3 \right)}{\left( x-2 \right)}$ , $x\ne 2,1$
Work Step by Step
Consider the expression,
$\frac{{{x}^{2}}+2x-3}{{{x}^{2}}-3x+2}$
The given expression can be further evaluated as
$\begin{align}
& \frac{{{x}^{2}}+2x-3}{{{x}^{2}}-3x+2}=\frac{{{x}^{2}}+3x-x-3}{{{x}^{2}}-2x-x+2} \\
& =\frac{x\left( x+3 \right)-1\left( x+3 \right)}{x\left( x-2 \right)-1\left( x-2 \right)} \\
& =\frac{\left( x+3 \right)\left( x-1 \right)}{\left( x-2 \right)\left( x-1 \right)} \\
& =\frac{\left( x+3 \right)}{\left( x-2 \right)}
\end{align}$
Here, $x\ne 2,1$ because at $x=2,1$ the denominator will become zero.
Therefore, the value of $\frac{{{x}^{2}}+2x-3}{{{x}^{2}}-3x+2}$ is $\frac{\left( x+3 \right)}{\left( x-2 \right)}$.
