Bài giải:

$\frac{4x-7}{{{x}^{2}}-3x+2}$ = $\frac{a}{x-1}$ + $\frac{b}{x-2}$

$\frac{4x-7}{{{x}^{2}}-3x+2}$= $\frac{a(x-2)+b(x-1)}{(x-1)(x-2)}$

$\frac{4x-7}{{{x}^{2}}-3x+2}$= $\frac{(a+b)x-(2a+b)}{{{x}^{2}}-3x+2}$

Suy ra: $\left\{ \begin{align}& 4=a+b \\ & 7=2a+b \\ \end{align} \right.$

$\Rightarrow$$\left\{ \begin{align}& 2a+b-a-b=7-4 \\ & a+b=4 \\ \end{align} \right.$

$\Rightarrow$$\left\{ \begin{align}& a=3 \\ & b=1 \\ \end{align} \right.$

Vậy đáp án đúng là: C