Question

[đặng minh ngọc]

Tìm số nguyên n sao cho n^2+5 chia hết cho n+1

Answer 1

[Kiệt Nguyễn]

\(\left(n^2+5\right)⋮\left(n+1\right)\)

\(\Rightarrow\left(n^2+n-n-1+6\right)⋮\left(n+1\right)\)

\(\Rightarrow\left[n\left(n+1\right)-\left(n+1\right)+6\right]⋮\left(n+1\right)\)

Vì \(\hept{\begin{cases}\left[n\left(n+1\right)\right]⋮\left(n+1\right)\\\left(n+1\right)⋮\left(n+1\right)\end{cases}}\)nên \(6⋮\left(n+1\right)\)

\(\Rightarrow n+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Lập bảng:

\(n+1\)	\(-1\)	\(1\)	\(-2\)	\(2\)	\(-3\)	\(3\)	\(-6\)	\(6\)
\(n\)	\(-2\)	\(0\)	\(-3\)	\(1\)	\(-4\)	\(2\)	\(-7\)	\(5\)

Vậy \(n\in\left\{-2;0;-3;1;-4;2;-7;5\right\}\)

Answer 2

[Nguyễn Hoàng Minh Thư]

Ta có:n^2+5\(⋮\)n+1   (1)

Mà n+1\(⋮\)n+1

\(\Rightarrow\)n.(n+1)\(⋮\)n+1

\(\Rightarrow\)n^2+n\(⋮\)n+1   (2)

Từ 1 và 2 

\(\Rightarrow\)n^2+n-n^2-5\(⋮\)n+1

\(\Rightarrow\)n-5\(⋮\)n+1

\(\Rightarrow\)n+1-6\(⋮\)n+1

\(\Rightarrow\)-6\(⋮\)n+1

\(\Rightarrow\)n+1\(\in\)Ư(-6)

\(\Rightarrow\)n+1\(\in\){1;-1;2;-2;3;-3;6;-6}

\(\Rightarrow\)n\(\in\){0;-2;1;-3;2;-4;5;-7}

Kết hợp điều kiện n\(\in\)\(ℤ\)nên n\(\in\){0;-2;1;-3;2;-4;5;-7}

Vậy n\(\in\){0;-2;1;-3;2;-4;5;-7}