Bài giải:

$\frac{{{x}^{2}}+3xy+2{{y}^{2}}}{{{x}^{3}}+2{{x}^{2}}y-x{{y}^{2}}-2{{y}^{3}}}$= $\frac{{{x}^{2}}+3xy+2{{y}^{2}}}{{{x}^{3}}-{{x}^{2}}y+3{{x}^{2}}y-3x{{y}^{2}}+2x{{y}^{2}}-2{{y}^{3}}}$

= $\frac{{{x}^{2}}+3xy+2{{y}^{2}}}{{{x}^{2}}(x-y)+3xy(x-y)+2{{y}^{2}}(x-y)}$ = $\frac{{{x}^{2}}+3xy+2{{y}^{2}}}{(x-y)({{x}^{2}}+3xy+2{{y}^{2}})}$

= $\frac{1}{x-y}$

Vậy đáp án đúng là: A