Để `(n^3-2n^2+3)/(n-2)` là số nguyên.
`\to n^3-2n^2+3\vdots n-2`
`\to n^2(n-2)+3\vdots n-2`
Vì `n^2(n-2)\vdots n-2`
`\to 3\vdots n-2`
`\to n-2\in Ư(3)=\{-3;-1;1;3\}`
\(→\left[ \begin{array}{l}n-2=-3\\n-2=-1\\n-2=1\\n-2=3\end{array} \right.\)
\(→\left[ \begin{array}{l}n=-1\\n=1\\n=3\\n=5\end{array} \right.\)
Vậy `n∈\{-1;1;3;5\}` để `(n^3-2n^2+3)/(n-2)` là số nguyên.