Ta có \(\left(n^2+7n+9\right)⋮\left(n+3\right)\)
\(\Leftrightarrow\left[\left(n^2+3n\right)+\left(4n+12\right)-3\right]⋮\left(n+3\right)\)
\(\Leftrightarrow\left[n\left(n+3\right)+4\left(n+3\right)-3\right]⋮\left(n+3\right)\)
\(\Rightarrow-3⋮\left(n+3\right)\)Hay \(n+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\)
| n + 3 | - 3 | - 1 | 1 | 3 |
| n | - 6 | - 4 | - 2 | 0 |
Vậy \(n\in\left\{-6;-4;-2;0\right\}\)
Ta có: \(\frac{n^2+7n+9}{n+3}=\frac{n^2+3n+3n+9}{n+3}+\frac{n}{n+3}\)
= \(\frac{\left(n+3\right)^2}{n+3}+\frac{n+3-3}{n+3}=n+3+1-\frac{3}{n+3}\)=> x + 4 - 3/n+3
Do n thuộc N => n+ 4 thuộc N; Để \(n^2+7n+9⋮n+3=>3⋮n+3\)
Hay n+3 thuộc Ư(3)
=> n+ 3 thuộc { -3;-1;1;3}
=> n thuộc { -6; -4; -2;0}
Mà n thuộc N nên n =0
