\(C=\dfrac{3x^3-x^2+6x+6}{3x-1}=\dfrac{x^2\left(3x-1\right)+2\left(3x-1\right)+8}{3x-1}=x^2+x+\dfrac{8}{3x-1}\)
Để C nguyên thì \(\left(3x-1\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
Do \(x\in Z\)
\(\Rightarrow x\in\left\{-1;0;1;3\right\}\)
\(D=\dfrac{2x^3+3x^2-x+4}{2x+1}=\dfrac{x^2\left(2x+1\right)+x\left(2x+1\right)-\left(2x+1\right)+5}{2x+1}=x^2+x-\dfrac{5}{2x+1}\)
Để D nguên thì:
\(\left(2x+1\right)\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(x\in Z\)
\(\Rightarrow x\in\left\{-3;-1;0;2\right\}\)
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