Ta có : \(4n-5=4\left(n-3\right)+7\)
Để \(n\in Z\) thì \(\left(4n-5\right)⋮n-3\\ \Rightarrow4\left(n-3\right)⋮n-3;7⋮n-3\\ \Rightarrow\left(n-3\right)\inƯ\left(7\right)\)
Mà \(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)
*Nếu \(n-3=-7\Rightarrow n=-4\)
*Nếu \(n-3=-1\Rightarrow n=2\)
*Nếu \(n-3=1\Rightarrow n=4\)
* Nếu \(n-3=7\Rightarrow n=10\)
Vậy \(n\in\left\{-4;2;4;10\right\}\)
Ta có: \(4n-5⋮n-3\)
\(\Leftrightarrow4n-12+7⋮n-3\)
Mà \(4n-12⋮n-3\)
\(\Leftrightarrow7⋮n-3\)
\(\Leftrightarrow n-3\inƯ_{\left(7\right)}=\left\{-7;-1;1;7\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}n-3=-7\\n-3=-1\\n-3=1\\n-3=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=-4\\2\\4\\10\end{matrix}\right.\)
Vậy: \(n=-4;2;4;10\)
