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Question

Tìm $n\in z$ để : ( n2 + 2n + 7 ) $⋮$(n + 1)

Vũ Mạnh Chí · Accepted Answer

$\frac{n^2+2n+7}{n+1}=\frac{\left(n^2+2n+1\right)+6}{n+1}=\frac{\left(n+1\right)^2+6}{n+1}=\left(n+1\right)+\frac{6}{n+1}$(n$\ne$-1)

Để $\left(n^2+2n+7\right)⋮\left(n+1\right)$ thì n+1$\in$Ư(6)

Ta có bảng sau:

n+1
			-6
			-1
			1
			6

n
			-7
			-2
			0
			5

Vậy n$\in${-7;-2;0;5}Câu trả lời: $\frac{n^2+2n+7}{n+1}=\frac{\left(n^2+2n+1\right)+6}{n+1}=\frac{\left(n+1\right)^2+6}{n+1}=\left(n+1\right)+\frac{6}{n+1}$(n$\ne$-1)

Để $\left(n^2+2n+7\right)⋮\left(n+1\right)$ thì n+1$\in$Ư(6)

Ta có bảng sau:

n+1
			-6
			-1
			1
			6

n
			-7
			-2
			0
			5

Vậy n$\in${-7;-2;0;5}

Trần Quỳnh Mai · Answer

Ta có : $n^2+2n+7⋮n+1\Rightarrow n^2+n+n+7⋮n+1$

$\Rightarrow n\left(n+1\right)+n+7⋮n+1$

Vì : $n\left(n+1\right)⋮n+1\Rightarrow n+7⋮n+1$

Mà : $n+1⋮n+1\Rightarrow\left(n+7\right)-\left(n+1\right)⋮n+1$

$\Rightarrow n+7-n-1⋮n+1\Rightarrow6⋮n+1$

$\Rightarrow n+1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}$

$\Rightarrow n\in\left\{0;-2;1;-3;2;-4;5;-7\right\}$

Vậy ...Câu trả lời: Ta có : $n^2+2n+7⋮n+1\Rightarrow n^2+n+n+7⋮n+1$