Bài giải:

$\frac{{{x}^{2}}+x}{x-1}.Q\left( x \right)=\frac{{{x}^{2}}+3x+2}{x\left( {{x}^{2}}-1 \right)}$

$\Rightarrow Q\left( x \right)=\frac{{{x}^{2}}+3x+2}{x\left( {{x}^{2}}-1 \right)}:\frac{{{x}^{2}}+x}{x-1}$

$=\frac{{{x}^{2}}+2x+x+2}{x\left( x-1 \right)\left( x+1 \right)}.\frac{x-1}{x\left( x+1 \right)}$

$=\frac{x\left( x+2 \right)+\left( x+2 \right)}{x\left( x-1 \right)\left( x+1 \right)}.\frac{x-1}{x\left( x+1 \right)}$

=$\frac{\left( x+2 \right)\left( x+1 \right)}{x\left( x-1 \right)\left( x+1 \right)}.\frac{x-1}{x\left( x+1 \right)}$

=$\frac{x+2}{{{x}^{3}}+{{x}^{2}}}$

Đáp án đúng là C