​$$\begin{gathered} A=(\frac{\mathrm{x}}{(\mathrm{x}+6)(\mathrm{x}-6)}-\frac{x-6}{x(x+6)}) : \frac{2(x-3)}{x(x+6)}+\frac{2x}{6-x} \\ =\frac{x^2-{(x-6)}^2}{x(x+6)(x-6)}.\frac{x(x+6)}{2(x-3)}+\frac{2x}{6-x} \\ =\frac{12x-36}{2(x-3)(x-6)}+\frac{2x}{6-x}=\frac{12(x-3)}{2(x-3)(x-6)}+\frac{2x}{6-x} \\ =\frac{6}{x-6}-\frac{2x}{x-6}=\frac{6-2x}{x-6} \end{gathered}$$


$$\begin{gathered} {\mathrm{x}}^2-\mathrm{x}-6=0 \iff (\mathrm{x}-3)(\mathrm{x}+2)=0\iff \mathrm{x}=3 \mathrm{hoặc} \mathrm{x}=-2 \\ \odot \mathrm{với} \mathrm{x}=3 \iff \mathrm{A}=\frac{6-2.3}{3-6}=0 \\ \odot \mathrm{với} \mathrm{x}=-2 \iff \mathrm{A}=\frac{6-2.(-2)}{-2-6}=\frac{-5}{4} \end{gathered}$$    (k bt ý của bài phải như vầy k?? :v)


$$\begin{gathered} \mathrm{để} \mathrm{A}>-2 \mathrm{thì}\frac{6-2\mathrm{x}}{\mathrm{x}-6}>-2 \\ \iff \frac{6-2\mathrm{x}}{\mathrm{x}-6}+2>0\iff \frac{6-2\mathrm{x}+2\mathrm{x}-12}{\mathrm{x}-6}>0 \\ \iff \frac{-6}{\mathrm{x}-6}>0\iff \mathrm{x}-6<0\iff \mathrm{x}<6 \end{gathered}$$


nếu có sai sót thì mình sẽ sửa lại nhé! làm cực lắm ấy! @@!