a: \(A=\left(\dfrac{x^2+2}{x^3-1}+\dfrac{x}{x^2+x+1}+\dfrac{1}{1-x}\right):\dfrac{x-1}{2}\)
\(=\dfrac{x^2+2+x^2-x-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{x^2-2x+1}{\left(x-1\right)^2}\cdot\dfrac{2}{x^2+x+1}=\dfrac{2}{x^2+x+1}\)
b: Để A=2 thì \(x^2+x+1=0\)
=>x(x+1)=0
=>x=0 hoặc x=-1