ĐKXĐ: x\(\ne3,x\ne-3\)
\(\Rightarrow\left(x-a\right)\left(a-3\right)+\left(x+3\right)\left(a+3\right)=-6a\)
\(\Leftrightarrow xa-3x-a^2+3a+ax+3x+3a+3=-6a\)
\(\Leftrightarrow2ax-a^2+12a+3=0\) \(\Leftrightarrow2ax=a^2-12a-3\Leftrightarrow x=\dfrac{a^2}{2}-6a-\dfrac{3}{2}\)(TM)
Vậy...