Với \(x\ne\frac{1}{2}\), \(x\in Z\), ta có:
\(P=\frac{7x-7x^2+2x^3+3}{2x-1}\)
\(=x^2-3x+2+\frac{5}{2x-1}\)
Để P đạt giá trị nguyên thì \(\frac{5}{2x-1}\) đạt giá trị nguyên (vì \(x^2-3x+2\in Z\)với mọi \(x\in Z\))
Hay \(5⋮\left(2x-1\right)\)
\(\Rightarrow2x-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
TH1: \(2x-1=\pm1\Leftrightarrow\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
TH2: \(2x-1=\pm5\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;-2;3\right\}\)
