bài 1
để A∈Z
\(=>n+3\inƯ\left(1\right)=\left\{-1;1\right\}\)
\(=>\left\{{}\begin{matrix}n+3=-1\\n+3=1\end{matrix}\right.=>\left\{{}\begin{matrix}n=-4\\n=-2\end{matrix}\right.\)
vậy \(n\in\left\{-4;-2\right\}\) thì \(A\in Z\)
Để A nguyên
⇒ \(\left(n+3\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)
n+3 1 -2
n -2 -4
\(B=\dfrac{n+3+1}{n+1}=1+\dfrac{3}{n+1}\)
Để B nguyên
\(\Rightarrow\left(n+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
n+1 1 -1 3 -3
n 0 -2 2 -4
bài 2
\(B=\dfrac{n+4}{n+1}=\dfrac{n+1}{n+1}+\dfrac{3}{n+1}=1+\dfrac{3}{n+1}\)
để \(B\in Z\Rightarrow n+1\inƯ\left(3\right)=\left\{-1;1;-3;3\right\}\)
\(=>\left\{{}\begin{matrix}n+1=-1\\n+1=1\\n+1=-3\\n+1=3\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}n=-2\\n=0\\n=-4\\n=2\end{matrix}\right.\)
vậy \(n\in\left\{-2;0;4;2\right\}\left(thì\right)B\in Z\)
\(B=\dfrac{n+4}{n+1}=\dfrac{n+1+3}{n+1}=\dfrac{3}{n+1}\\ \Rightarrow3⋮n+1\\ \Rightarrow n+1\in\text{Ư}\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng :
| n+1 | 1 | -1 | 3 | -3 |
| n | 0 | -2 | 2 | -4 |
