Question

Tìm n \(\in\) N*

\(n^2-3\) \(\vdots\) \(n+3\)

 

Answer 1

Ta có:\(n^2-3⋮n+3\)

\(\Leftrightarrow n^2+3n-3n-9+6⋮n+3\)

\(\Leftrightarrow\left(n^2+3n\right)-\left(3n+9\right)+6⋮n+3\)

\(\Leftrightarrow n\left(n+3\right)-3\left(n+3\right)+6⋮n+3\)

\(\Leftrightarrow6⋮n+3\)

\(\Leftrightarrow n+3\inƯ\left(6\right)\)

Mà \(n\in N\)*\(\Rightarrow n+3\ge4\)

\(\Leftrightarrow n+3=6\)

\(\Leftrightarrow n=3\)

 

 

 

 

Answer 2

\(n^2-3⋮n+3\\ \Rightarrow\left(n-3\right)\left(n+3\right)+6⋮n+3\\ \Rightarrow6⋮n+3\Rightarrow n+3\in\text{Ư}\left(6\right)\)

Tới đây dễ rồi nha!