Ta có : \(2n-1⋮n-3\)
\(\Leftrightarrow2n-6+5⋮n-3\)
Thấy \(2n-6=2\left(n-3\right)⋮n-3\)
\(\Rightarrow5⋮n-3\)
- Để 5 chia hết cho n - 3 <=> \(n-3\inƯ_{\left(5\right)}\)
\(\Leftrightarrow n-3\in\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow n\in\left\{4;2;8;-2\right\}\)
Vậy ...
\(2n-1⋮n-3\)
\(\Leftrightarrow2n-6+5⋮n-3\)
\(Vì:2n-6⋮n-3\left(n\in R\right)\)
\(\Rightarrow5⋮n-3\)
\(\Rightarrow n-3\inƯ\left(5\right)=\left\{-1,1,-5,5\right\}\)
\(\Rightarrow n\in\left\{2,4,-2,8\right\}\)
Ta có: \(2n-1⋮n-3\)
\(\Leftrightarrow2n-6+5⋮n-3\)
mà \(2n-6⋮n-3\)
nên \(5⋮n-3\)
\(\Leftrightarrow n-3\inƯ\left(5\right)\)
\(\Leftrightarrow n-3\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{4;2;8;-2\right\}\)
Vậy: \(n\in\left\{4;2;8;-2\right\}\)
