a: ĐKXĐ: a<>1
b: \(A=\dfrac{4a^2-3a+17}{a^3-1}+\dfrac{2a-1}{a^2+a+1}+\dfrac{6}{1-a}\)
\(=\dfrac{4a^2-3a+17+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-12a+12}{\left(a-1\right)\left(a^2+a+1\right)}=\dfrac{-12}{a^2+a+1}\)
c: Để A là số nguyên âm thì \(-12⋮a^2+a+1\)
\(\Leftrightarrow a^2+a+1\in\left\{1;2;3;4;6;12\right\}\)
\(\Leftrightarrow a^2+a+1\in\left\{1;3\right\}\)
\(\Leftrightarrow a\in\left\{-2;0\right\}\)