Để \(\frac{3n+1}{5-2n}\in Z\) thì \(3n+1\) ⋮ \(5-2n\)
\(\Rightarrow\) \(2\left(3n+1\right)+3\left(5-2n\right)\)⋮ \(5-2n\)
(Vì \(3\left(5-2n\right)\) ⋮ \(5-2n\) ; \(2\left(3n+1\right)\) ⋮ \(5-2n\) )
\(\Rightarrow\) \(6n+2+15-6n\) ⋮ \(5-2n\)
\(\Rightarrow17\) ⋮ \(5-2n\)
\(\Rightarrow5-2n\in\text{Ư}\left(17\right)=\left\{1;17;-1;-17\right\}\)
\(\Rightarrow2n\in\left\{4;-12;6;22\right\}\)
\(\Rightarrow n\in\left\{2;-6;3;11\right\}\)
Vậy để phân số \(\frac{3n+1}{5-2n}\in Z\) thì \(n\in\left\{2;-6;3;11\right\}\)