Bài giải:

$\begin{align} & \frac{x}{{{x}^{2}}-4}+\frac{3}{{{\left( x+2 \right)}^{2}}} \\ & =\frac{x}{\left( x-2 \right)\left( x+2 \right)}+\frac{3}{{{\left( x+2 \right)}^{2}}} \\& =\frac{x\left( x+2 \right)+3\left( x-2 \right)}{\left( x-2 \right){{\left( x+2 \right)}^{2}}} \\ & =\frac{{{x}^{2}}+2x+3x-6}{\left( x-2 \right){{\left( x+2 \right)}^{2}}} \\& =\frac{{{x}^{2}}-x+6x-6}{\left( x-2 \right){{\left( x+2 \right)}^{2}}} \\& =\frac{\left( x-1 \right)\left( x+6 \right)}{\left( x-2 \right){{\left( x+2 \right)}^{2}}} \\\end{align}$

$\frac{\left( x-1 \right)\left( x+6 \right)}{\left( x-2 \right){{\left( x+2 \right)}^{2}}}=0$  khi (x-1)(x+6)=0 và (x-2­)(x+2)$\ne$ 0

Vậy x=1;x=-6

Đáp án đúng là C