\(n-3\text{⋮}n-3\)
\(\Rightarrow n\left(n-3\right)\text{⋮}n-3\)
\(\Rightarrow n^2-3n\text{⋮}n-3\)
Mà \(n^2-3\text{⋮}n-3\)
\(\Rightarrow\left(n^2-3\right)-\left(n^2-3n\right)\text{⋮}n-3\)
\(\Rightarrow3n-3\text{⋮}n-3\)
Lại có:\(n-3\text{⋮}3\Rightarrow3\left(n-3\right)\text{⋮}n-3\)
\(\Rightarrow3n-9\text{⋮}3\)
Mà \(3n-3\text{⋮}n-3\)
\(\Rightarrow\left(3n-3\right)-\left(3n-9\right)\text{⋮}n-3\)
\(6\text{⋮}n-3\)
\(\Rightarrow n-3\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow n\in\left\{-3;0;1;2;4;5;6;9\right\}\)
Vậy \(n\in\left\{-3;0;1;2;4;5;6;9\right\}\)
Ta có :
n2 - 3 chia hết cho n - 3
<=> n.n - 3 chia hết cho n -3
<=> 2n - 3 chia hết cho n - 3
<=> 2n - 6 + 3 chia hết cho n - 3
<=> 2.(n-3) + 3 chia hết cho n - 3
<=> 3 chia hết cho - 3
<=> n - 3 \(\in\) Ư(3) = {\(\pm\)1;\(\pm\)3}
<=> \(\begin{cases}n-3=1\Rightarrow n=4\\n-3=-1\Rightarrow n=2\\n-3=3\Rightarrow n=6\\n-3=-3\Rightarrow n=0\end{cases}\)
Vây n \(\in\) {4;2;6;0}