=>\(\dfrac{3n^3+10n^2-5}{3n+1}=\) \(\left(n^2+3n-1\right)-\dfrac{4}{3n+1}\)
để\(\dfrac{3n^3+10n^2-5}{3n+1}\) nguyên thì -\(\dfrac{4}{3n+1}\) nguyên
=>\(-4⋮\left(3n+1\right)\)
=>(3n+1)\(\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
ta có bảng sau
| 3n+1 | -1 | 1 | -2 | 2 | -4 | 4 |
| 3n | -2 | 0 | -3 | 2 | -5 | 3 |
| n | \(\dfrac{-2}{3}\) | 0 | -1 | \(\dfrac{2}{3}\) | \(\dfrac{-5}{3}\) | 1 |
mà n \(\in Z\)
=>n\(\in\) {-1;0;1}
vậy ....
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