Answer
The required value is $\frac{11}{\left( x-3 \right)\left( x-4 \right)}$ , $x\ne 3,4$.
Work Step by Step
Consider the expression,
$\frac{2x+3}{{{x}^{2}}-7x+12}-\frac{2}{x-3}$
The given expression can be evaluated as:
$\begin{align}
& \frac{2x+3}{{{x}^{2}}-7x+12}-\frac{2}{x-3}=\frac{2x+3}{{{x}^{2}}-4x-3x+12}-\frac{2}{x-3} \\
& =\frac{2x+3}{x\left( x-4 \right)-3\left( x-4 \right)}-\frac{2}{x-3} \\
& =\frac{2x+3}{\left( x-3 \right)\left( x-4 \right)}-\frac{2}{\left( x-3 \right)}
\end{align}$
Take the least common denominator:
$\begin{align}
& \frac{2x+3}{{{x}^{2}}-7x+12}-\frac{2}{x-3}=\frac{2x+3}{\left( x-3 \right)\left( x-4 \right)}-\frac{2\left( x-4 \right)}{\left( x-3 \right)\left( x-4 \right)} \\
& =\frac{2x+3-2x+8}{\left( x-3 \right)\left( x-4 \right)} \\
& =\frac{11}{\left( x-3 \right)\left( x-4 \right)}
\end{align}$
Here, $x\ne 3,4$ because at $x=3,4$ the denominator will become zero.
Therefore, the value of $\frac{2x+3}{{{x}^{2}}-7x+12}-\frac{2}{x-3}$ is $\frac{11}{\left( x-3 \right)\left( x-4 \right)}$.
