Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
tìm n thuộc Z
a, n2 +3n -13 chia hết n + 3
\(n^2+3n-13⋮n+3\)
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 ..
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
\(n^2+3n-13⋮n+3\)
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 ..
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 ..