Đáp án:
`D = (3a + 2)/(2a - 1)`
`text{Để D nguyên}`
`-> 3a + 2 \vdots 2a - 1`
`-> 2 (3a + 2) \vdots 2a - 1`
`-> 6a + 4 \vdots 2a - 1`
`-> 3 (2a - 1) + 7 \vdots 2a - 1`
`text{Vì}` `3 (2a - 1) \vdots 2a - 1`
`-> 7 \vdots 2a - 1`
`-> 2a- 1 ∈ Ư (7) = {±1; ±7) (a ∈ ZZ)`
`text{Ta có bảng :}`
$\begin{array}{|c|c|c|c|c|c|c|}\hline 2a - 1& 1 & -1 & 7 & -7 \\\hline a& 1 & 0 & 4 & -3 \\\hline\end{array}$
`text{Vậy a = (1;0;4;-3) để D nguyên}`
