a) Ta thực hiện phép chia \(3x^3+13x^2-7x+5\) cho \(3x-2\). Khi đó ta có:
\(A=\frac{3x^3+13x^2-7x+5}{3x-2}=3x^2+5x+1+\frac{7}{3x-2}\)
Nếu x nguyên thì \(3x^2+5x+1\in\text{Z}\) nên để A nguyên thì \(\frac{7}{3x-2}\in Z\)
\(\Rightarrow3x-2\in\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x\in\left\{1;3\right\}\)
b) Ta có: \(B=\frac{2x^5+4x^4-7x^3-44}{2x^2-7}=\left(x^3+2x^2+7\right)+\frac{5}{2x^2-7}\)
Để B nguyên thì \(\frac{5}{2x^2-7}\in Z\Rightarrow2x^2-7\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-1;1;2;-2\right\}\)