Bài $1$.
Ta có : $\dfrac{n+15}{n+3} = \dfrac{n+3+12}{n+3} = 1 + \dfrac{12}{n+3}$
Để $\dfrac{n+15}{n+3}$ là số tự nhiên thì : $12 \vdots n+3$
$⇒ n+3∈$ `Ư(12)={1;2;3;4;6;12}`
Do $n$ $∈$ $N$ $⇒$ $n+3$ $∈$ `{3;4;6;12}`
$⇔ n ∈$ `{0;1;3;9}`
Vậy $n ∈$ `{0;1;3;9}`.
Bài $2$.
$3n+4 \vdots n-1$
$⇔ 3n+4 - 3.(n-1) \vdots n-1$
$⇔ 3n+4 - 3n + 3 \vdots n-1$
$⇔ 7 \vdots n-1$
$⇒ n-1$ $∈$ `Ư(7)={1;7}`
$⇒ n$ $∈$ `{2;8}`
Vậy $n$ $∈$ `{2;8}`.
Bài $3$.
$4n-5 \vdots 2n-1$
$⇔ 4n-5 - 2.(2n-1) \vdots 2n-1$
$⇔ 4n-5 -4n + 2 \vdots 2n-1$
$⇔ -3 \vdots 2n-1$
$⇒ 2n-1$ $∈$ `Ư(3)={1;3}`
$⇒ n$ $∈$ `{1;2}`
Vậy $n$ $∈$ `{1;2}`.
Bài $4$.
$n^2 + 3n + 6 \vdots n+3$
$⇔ n(n+3) + 6 \vdots n+3$
$⇒ 6 \vdots n+3$
$⇒ n+3$ $∈$ `Ư(6)={1;2;3;6}`
Mà $n$ $∈$ $N$ $⇒$ $n+3$ $∈$ `{3;6}`
$⇒ n$ $∈$ `{0;3}`
Vậy $n$ $∈$ `{0;3}`.