a)Ta có \(M=\left(\dfrac{x-2}{x+1}+\dfrac{x+2}{1-x}+\dfrac{6x}{x^2-1}\right).\dfrac{x^3-1}{x-1}=\left(\dfrac{x-2}{x+1}-\dfrac{x+2}{x-1}+\dfrac{6x}{x^2-1}\right).\left(x^2+x+1\right)=\left(\dfrac{\left(x-2\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)}{x^2-1}+\dfrac{6x}{x^2-1}\right).\left(x^2+x+1\right)=0.\)b)Ta có \(2x^2=M=0\Leftrightarrow x=0\left(tm\right)\)