Bài giải:

$\frac{3{{x}^{2}}+5x+1}{{{x}^{3}}-1}-\frac{1-x}{{{x}^{2}}+x+1}-\frac{3}{x-1}$= $\frac{3{{x}^{2}}+5x+1}{(x-1)({{x}^{2}}+x+1)}-\frac{1-x}{{{x}^{2}}+x+1}-\frac{3}{x-1}$

= $\frac{3{{x}^{2}}+5x+1}{(x-1)({{x}^{2}}+x+1)}-\frac{(1-x)(x-1)}{(x-1)({{x}^{2}}+x+1)}-\frac{3({{x}^{2}}+x+1)}{(x-1)({{x}^{2}}+x+1)}$

= $\frac{3{{x}^{2}}+5x+1+{{x}^{2}}-2x+1-3{{x}^{2}}-3x-3}{(x-1)({{x}^{2}}+x+1)}$ = $\frac{{{x}^{2}}-1}{(x-1)({{x}^{2}}+x+1)}$

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

Vậy đáp án đúng là: A.