$\text{Đáp án:}$ $\text{ĐKXĐ:x\neq±3}$
$$\dfrac{x^2+2x+1}{(x-3)(x+3)}=6$$
$$⇔\dfrac{(x+1)^2}{x^2-9}=6$$
$$⇔(x+1)^2=6x^2-54$$
$$⇔x^2+2x+1-6x^2+54=0$$
$$⇔-5x^2+2x+55=0$$
$$⇔-5(x^2-\dfrac{2}{5}x)+55=0$$
$$⇔-5(x^2-\dfrac{2}{5}x+(\dfrac{1}{5})^2-(\dfrac{1}{5})^2)+55=0$$
$$⇔-5(x-\dfrac{1}{5})^2+\dfrac{276}{5}=0$$
$$⇔(x-\dfrac{1}{5})^2=\dfrac{276}{25}$$
$$⇔\left[ \begin{array}{l}x-\dfrac{1}{5}=\sqrt{\dfrac{276}{25}}\\ x-\dfrac{1}{5}=-\sqrt{\dfrac{276}{25}} \end{array} \right.$$
$$⇔\left[ \begin{array}{l}x=\sqrt{\dfrac{276}{25}}+\dfrac{1}{5}\\ x=-\sqrt{\dfrac{276}{25}}+\dfrac{1}{5} \end{array} \right.$$
$\text{Vậy............}$