Question

Tìm số nguyên n sao cho 2n- 1 chia hết cho n – 3

Answer 1

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 ...

Answer 2

\(2n-1⋮n-3\)

\(\Leftrightarrow2n-6+5⋮n-3\)

\(Vì:2n-6⋮n-3\left(n\in R\right)\)

\(\Rightarrow5⋮n-3\)

\(\Rightarrow n-3\inƯ\left(5\right)=\left\{-1,1,-5,5\right\}\)

\(\Rightarrow n\in\left\{2,4,-2,8\right\}\)

Answer 3

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: \(n\in\left\{4;2;8;-2\right\}\)