Question

[Lê Thị Hà Linh]

Tìm n để: n^2-3 chia hết cho n-3

Answer 1

[Lovers]

\(n-3\text{⋮}n-3\)

\(\Rightarrow n\left(n-3\right)\text{⋮}n-3\)

\(\Rightarrow n^2-3n\text{⋮}n-3\)

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 \(n\in\left\{-3;0;1;2;4;5;6;9\right\}\)

Answer 2

[Dũng Nguyễn Đình]

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}