Bài giải:

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

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

Suy ra: $\frac{{{({{x}^{2}}+2)}^{2}}-4{{x}^{2}}}{y({{x}^{2}}+1)-2xy-({{x}^{2}}-1)-1}\ne \frac{{{x}^{2}}+x+2}{y-1}$

Vậy đáp án đúng là: B