a: \(A=\left(\dfrac{2x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right):\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
\(=\dfrac{2x^2+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}:\dfrac{x^2+x+1-x^2+2}{x^2+x+1}\)
\(=\dfrac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+3}\)
\(=\dfrac{x}{x+3}\)
b: |x-2|=3
=>x-2=3 hoặc x-2=-3
=>x=5(nhận) hoặc x=-1(nhận)
Khi x=5 thì \(A=\dfrac{5}{5+3}=\dfrac{5}{8}\)
Khi x=-1 thì \(A=\dfrac{-1}{-1+3}=\dfrac{-1}{2}\)