\(A=\dfrac{3n+1}{n-4}\) ; \(\left(n\ne4\right)\)
\(A=\dfrac{3\left(n-4\right)+13}{n-4}\)
\(A=3+\dfrac{13}{n-4}\)
Để A nguyên thì \(\dfrac{13}{n-4}\in Z\Rightarrow n-4\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
`@n-4=1->n=5`
`@n-4=-1->n=3`
`@n-4=13->n=17`
`@n-4=-13->n=-9`
Vậy \(n\in\left\{5;3;17;-9\right\}\) thì A nguyên