Bài giải:

A – B = $\left( 2{{x}^{3}}-4{{x}^{2}}y+1\frac{1}{3}.x{{y}^{2}}-{{y}^{4}}+1 \right)-\left( -2{{x}^{3}}-1\frac{1}{2}.{{x}^{2}}y+\frac{2}{3}x{{y}^{2}}-{{y}^{4}}-3 \right)$

= $2{{x}^{3}}-4{{x}^{2}}y+1\frac{1}{3}x{{y}^{2}}-{{y}^{4}}+1+2{{x}^{3}}+1\frac{1}{2}{{x}^{2}}y-\frac{2}{3}x{{y}^{2}}+{{y}^{4}}+3$

= $2{{x}^{3}}-4{{x}^{2}}y+\frac{4}{3}x{{y}^{2}}-{{y}^{4}}+1+2{{x}^{3}}+\frac{3}{2}{{x}^{2}}y-\frac{2}{3}x{{y}^{2}}+{{y}^{4}}+3$

= $(2{{x}^{3}}+2{{x}^{3}})+\left( -4{{x}^{2}}y+\frac{3}{2}{{x}^{2}}y \right)+\left( \frac{4}{3}x{{y}^{2}}-\frac{2}{3}x{{y}^{2}} \right)+(-{{y}^{4}}+{{y}^{4}})+(1+3)$

= $4{{x}^{3}}-\frac{5}{2}{{x}^{2}}y+\frac{2}{3}x{{y}^{2}}+4$

Đáp án đúng là: C