ĐKXĐ: \(x\notin\left\{-2;3;-1\right\}\)\(\dfrac{x}{x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{2}{6-2x}\)=>\(\dfrac{x}{x+2}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{-1}{x-3}\)=>\(\dfrac{x}{x+2}=\dfrac{-1}{x-3}+\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{-x-1+2x}{\left(x-3\right)\left(x+1\right)}\)=>\(\dfrac{x}{x+2}=\dfrac{x-1}{\left(x-3\right)\left(x+1\right)}\)=>\(x\left(x^2-2x-3\right)=\left(x-1\right)\left(x+2\right)\)=>\(x^3-2x^2-3x-\left(x^2+x-2\right)=0\)=>\(x^3-3x^2-4x+2=0\)=>\(\left[{}\begin{matrix}x\simeq3,9\left(nhận\right)\\x\simeq0,4\left(nhận\right)\\x\simeq-1,3\left(nhận\right)\end{matrix}\right.\)Câu trả lời: ĐKXĐ: \(x\notin\left\{-2;3;-1\right\}\)\(\dfrac{x}{x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{2}{6-2x}\)=>\(\dfrac{x}{x+2}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{-1}{x-3}\)=>\(\dfrac{x}{x+2}=\dfrac{-1}{x-3}+\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{-x-1+2x}{\left(x-3\right)\left(x+1\right)}\)=>\(\dfrac{x}{x+2}=\dfrac{x-1}{\left(x-3\right)\left(x+1\right)}\)=>\(x\left(x^2-2x-3\right)=\left(x-1\right)\left(x+2\right)\)=>\(x^3-2x^2-3x-\left(x^2+x-2\right)=0\)=>\(x^3-3x^2-4x+2=0\)=>\(\left[{}\begin{matrix}x\simeq3,9\left(nhận\right)\\x\simeq0,4\left(nhận\right)\\x\simeq-1,3\left(nhận\right)\end{matrix}\right.\)