Ta có:
\(4n^2+n+2⋮n+1\)
\(\Rightarrow\left(4n^2+4n\right)-3n+2⋮n+1\)
\(\Rightarrow4n\left(n+1\right)-3n+2⋮n+1\)
\(\Rightarrow-3n+2⋮n+1\)
\(\Rightarrow-\left(3n+3\right)+5⋮n+1\)
\(\Rightarrow-3\left(n+1\right)+5⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in\left\{1;5\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=5\Rightarrow n=4\end{matrix}\right.\)
Vậy \(n\in\left\{0;4\right\}\)
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