ĐKXĐ: \(x\ne-\dfrac{1}{2}\)
Ta có: \(D=\dfrac{2a^3+a^2+2a+4}{2a+1}=\dfrac{a^2\left(2a+1\right)+\left(2a+1\right)+3}{2a+1}\)
\(=\dfrac{\left(2a+1\right)\left(a^2+1\right)+3}{2a+1}=\dfrac{\left(2a+1\right)\left(a^2+1\right)}{2a+1}+\dfrac{3}{2a+1}\) \(=a^2+1+\dfrac{3}{2a+1}\)
Để \(D\in Z\) <=> \(a^2+1+\dfrac{3}{2a+1}\in Z\)
=> \(\left\{{}\begin{matrix}a^2\in Z\\\dfrac{3}{2a+1}\in Z\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a\in Z\\\dfrac{3}{2a+1}\in Z\end{matrix}\right.\)
Để \(\dfrac{3}{2a+1}\in Z\) <=> \(3⋮2a+1\)
mà \(a\in Z\) => \(2a+1\inƯ_{\left(3\right)}=\left\{\pm1;\pm3\right\}\)
Ta có bảng:
| 2a+1 | 1 | -1 | 3 | -3 |
| a | 0 | -1 | 1 | -2 |
Vậy \(D\in Z\) khi \(a\in\left\{0;\pm1;-2\right\}\)
