2n +1 chia hết cho n -3
\(\Rightarrow2\left(n-3\right)+7⋮\)\(n-3\)
=> \(7⋮n-3\)
\(\Rightarrow n-3\inƯ\left(7\right)\)\(=\left\{\pm1;\pm7\right\}\)
Bạn xét từng trường hợp rồi tìm n nhé
Có \(\left(2n+1\right)⋮\left(n-3\right)\)
\(\Rightarrow2\left(n-3\right)+7⋮\left(n-3\right)\)
Mà \(2\left(n-3\right)⋮\left(n-3\right)\Rightarrow7⋮\left(n-3\right)\)
\(\Rightarrow n-3\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)
\(\Rightarrow n\in\left\{4;2;10;-4\right\}\)
Vậy n \(\in\left\{4;2;10;-4\right\}\)
\(\left(2n+1\right)⋮\left(n-3\right)\)
\(\Rightarrow2\left(n-3\right)+7⋮\left(n-3\right)\)
Mà \(2\left(n-3\right)⋮\left(n-3\right)=7⋮\left(n-3\right)\)
\(\Rightarrow n-3\in\text{Ư}\left(7\right)=\left\{1;-1;7;-7\right\}\)
\(\Rightarrow n\in\left\{4;2;10;-4\right\}\)
Vậy \(n\in\left\{4;2;10;-4\right\}\)
ban nao tra loi tot minh se h cho cac ban nhe
