Question

Cho A= \(n^3+2n^2-3n+2\)
B= \(n^2-n\)
Tìm \(n\in Z\text{ để }A\text{ }⋮\text{ }B\)

Answer 1

\(C=\dfrac{A}{B}=\dfrac{n^3+2n^2-3n+2}{n^2-n}=\dfrac{\left(n^3-n^2\right)+3n^2-3n+2}{n^2-n}=\dfrac{n\left(n^2-n\right)+3\left(n^2-n\right)+2}{n^2-n}\)\(C=n+3+\dfrac{2}{n^2-n}\)

\(n,C\in Z\Rightarrow\dfrac{2}{n^2-n}\in Z\Rightarrow n^2-n=\left\{-2;-1;1;2\right\}\)

n^2 -n là hai số chẵn

\(\left[{}\begin{matrix}n^2-n=-2\\n^2-n=2\end{matrix}\right.\)

\(\left[{}\begin{matrix}n^2-n=-2\left(vn\right)\\n^2-n=2\left[{}\begin{matrix}n_1=-1\\n_2=2\end{matrix}\right.\end{matrix}\right.\)