a: ĐKXĐ: a<>1
b: \(A=\dfrac{4a^2-3a+17+\left(2a-1\right)\left(a-1\right)-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-2a^2-9a+11+2a^2-3a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-12a+12}{\left(a-1\right)\left(a^2+a+1\right)}=\dfrac{-12}{a^2+a+1}\)