Vì \(n\in Z\Rightarrow\frac{3n-5}{n-4}\in Z\Rightarrow\left(3n-5\right)⋮\left(n-4\right)\)
Ta có:
\(3n-5=3n-12-5+12=3n-12+7\)
\(\Rightarrow3n-12+7⋮n-4\Rightarrow7⋮n-4\Rightarrow n-4=7\Rightarrow n-4\inƯ\left(7\right)\)
\(\Rightarrow n-4\in\){1;7}
\(\Rightarrow n\in\){5;11}