Question

n^2 - 9 + 37 ⋮ n-3

Answer 1

Ta có: \(n^2-9+37⋮n-3\)

\(\Leftrightarrow n-3\in\left\{1;-1;37;-37\right\}\)

hay \(n\in\left\{4;2;40;-34\right\}\)

Answer 2

\(n^2-9+37⋮n-3\\ \Rightarrow\left(n-3\right)\left(n+3\right)+37⋮n-3\)

Vì \(\left(n-3\right)\left(n+3\right)⋮\left(n-3\right)\\ \Rightarrow37⋮\left(n-3\right)\\ \Rightarrow n-3\inƯ\left(37\right)=\left\{\pm1;\pm37\right\}\\ \Rightarrow n\in\left\{-34;2;4;40\right\}\)