Bài giải:

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

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

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

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

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

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

Đáp án đúng là A