\(\dfrac{2}{x^2-x-6}+\dfrac{x+1}{x^2+x-12}=\dfrac{x}{x^2+6x+8}\)
\(\Leftrightarrow\dfrac{2}{\left(x-3\right)\left(x+2\right)}+\dfrac{x+1}{\left(x-3\right)\left(x+4\right)}=\dfrac{x}{\left(x+2\right)\left(x+4\right)}\)
=> 2(x+4)+(x+1)(x+2)=x(x-3)
⇔2x+8+x2+2x+x+2=x2-3x
⇔x2+5x+10=x2-3x
⇔x2-x2+5x+3x=-10
⇔8x=-10
\(\Leftrightarrow\dfrac{-5}{4}\)
Vậy S={-\(\dfrac{5}{4}\)}