Question

[Dương Bảo Lưu]

tìm x thuộc z để a=2n+7/n+1 thuộc z

Answer 1

[Huỳnh Bá Nhật Minh]

Để \(a=\frac{2n+7}{n+1}\inℤ\)thì \(2n+7⋮n+1\)

\(\Rightarrow\left(2n+7\right)⋮n+1\)

\(\Rightarrow\left(n+1\right)⋮n+1=\left(n+1\right)\cdot2⋮n+1=\left(2n+2\right)⋮n+1\)

\(\Rightarrow\left(2n+7\right)-\left(2n+2\right)⋮n+1\)

\(\Rightarrow2n+7-2n-2⋮n-1\)

\(\Rightarrow5⋮n-1\)

\(\Rightarrow n-1\inƯ\left(5\right)\)

\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta có bảng sau :

\(n+1\)	\(1\)	\(-1\)	\(5\)	\(-5\)
\(n\)	\(0\)	\(-2\)	\(4\)	\(-6\)

Vậy \(n\in\left\{0,-2,4,-6\right\}\)thì \(a\inℤ\)