Câu hỏi của Linh Trần

Question

tìm n thuộc Z

a, n2 +3n -13 chia hết n + 3

Nguyễn Thanh Hằng · Accepted Answer

$n^2+3n-13⋮n+3$

Mà $n+3⋮n+3$

$\Leftrightarrow\left\{{}\begin{matrix}n^2+3n-13⋮n+3\n^2+3n⋮n+3\end{matrix}\right.$

$\Leftrightarrow13⋮n+3$

$\Leftrightarrow n+3\inƯ\left(13\right)$

$\Leftrightarrow\left[{}\begin{matrix}n+3=1\n+3=13\n+3=-1\n+3=-13\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}n=-2\n=10\n=-4\n=-16\end{matrix}\right.$

Vậy ..Câu trả lời: $n^2+3n-13⋮n+3$

Mà $n+3⋮n+3$

$\Leftrightarrow\left\{{}\begin{matrix}n^2+3n-13⋮n+3\n^2+3n⋮n+3\end{matrix}\right.$

$\Leftrightarrow13⋮n+3$

$\Leftrightarrow n+3\inƯ\left(13\right)$

$\Leftrightarrow\left[{}\begin{matrix}n+3=1\n+3=13\n+3=-1\n+3=-13\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}n=-2\n=10\n=-4\n=-16\end{matrix}\right.$

Vậy ..

Diệu Thảo Nguyễn Hà · Answer

n2+3n−13⋮n+3n2+3n−13⋮n+3

Mà n+3⋮n+3n+3⋮n+3

⇔⎧⎨⎩n2+3n−13⋮n+3n2+3n⋮n+3⇔{n2+3n−13⋮n+3n2+3n⋮n+3

⇔13⋮n+3⇔13⋮n+3

⇔n+3∈Ư(13)⇔n+3∈Ư(13)

⇔⎡⎢ ⎢ ⎢⎣n+3=1n+3=13n+3=−1n+3=−13⇔[n+3=1n+3=13n+3=−1n+3=−13

⇔⎡⎢ ⎢ ⎢⎣n=−2n=10n=−4n=−16⇔[n=−2n=10n=−4n=−16

Vậy ..Câu trả lời: n2+3n−13⋮n+3n2+3n−13⋮n+3