Bài giải:

Thay x = $\frac{1}{2}$ vào phương trình ta có:

$\frac{\frac{1}{2}+a}{a-\frac{1}{2}}+\frac{\frac{1}{2}-a}{a+\frac{1}{2}}=\frac{a(3a+1)}{{{a}^{2}}-\frac{1}{4}}$

ĐKXĐ: a $\ne $$\frac{1}{2}$ ; a $\ne $$-\frac{1}{2}$

$\Leftrightarrow \frac{\frac{1+2a}{2}}{\frac{2a-1}{2}}+\frac{\frac{1-2a}{2}}{\frac{2a+1}{2}}=\frac{a(3a+1)}{\left( a-\frac{1}{2} \right)\left( a+\frac{1}{2} \right)}$

$\Leftrightarrow \frac{1+2a}{2a-1}+\frac{1-2a}{2a+1}=\frac{4a(3a+1)}{(2a-1)(2a+1)}$

$\Leftrightarrow \frac{{{(1+2a)}^{2}}}{(1+2a)(2a-1)}+\frac{(1-2a)(2a-1)}{(2a+1)(2a-1)}=\frac{4a(3a+1)}{(2a-1)(2a+1)}$

$\Leftrightarrow $1 + 4a + 4a² – 4a² + 4a – 1 = 12a² + 4a

$\Leftrightarrow $8a = 12a² + 4a

$\Leftrightarrow $12a² – 4a = 0

$\Leftrightarrow $4a(3a – 1) = 0

$\Leftrightarrow \left[ \begin{align}& a=0 \\  & a=\frac{1}{3} \\ \end{align} \right.$           (thỏa mãn)

Vậy đáp án đúng là: C