Vì: \(\frac{13}{n-1}.\frac{n}{3}\inℤ\)( \(n\ne1\))
\(\Rightarrow\frac{13n}{3.\left(n-1\right)}=\frac{13n}{3n-3}\inℤ\)
\(\Rightarrow13n⋮3n-3\)
\(\Rightarrow4.\left(3n-3\right)+n+12⋮3n-3\)
\(\Rightarrow n+12⋮3n-3\)
\(\Rightarrow3.\left(n-12\right)⋮3n-3\)
\(\Rightarrow3n-36⋮3n-3\)
\(\Rightarrow\left(3n-2\right)-33⋮3n-2\)
\(\Rightarrow33⋮n-2\)
\(\Rightarrow n-2\inƯ\left(33\right)=\left\{-33;-11;-3;-1;1;3;11;33\right\}\)
\(\Rightarrow n\in\left\{-31;-9;-1;1;3;5;13;35\right\}\)
Vậy: .......