Question

[Đẹp Trai Không Bao Giờ Sai]

Tìm n để

\(n^2+2n-4⋮11\)

\(n^4-2n^3+2n^2-2n+1⋮n^4+1\)

\(n^3-n^2+2n+7⋮n^2+1\)

Answer 1

[bảo nam trần]

a, \(n^2+2n-4=n^2+2n-15+11=\left(n-3\right)\left(n-5\right)+11\)

Để \(n^2+2n-4⋮11\Leftrightarrow\left(n-3\right)\left(n+5\right)⋮11\Leftrightarrow\left[{}\begin{matrix}n-3⋮11\\n+5⋮11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=BS11+3\\n=BS11-5\end{matrix}\right.\)

c,\(\dfrac{n^3-n^2+2n+7}{n^2+1}=\dfrac{n^3+n-n^2-1+n+8}{n^2+1}=\dfrac{n\left(n^2+1\right)-\left(n^2+1\right)+n+8}{n^2+1}=n-1+\dfrac{n+8}{n^2+1}\)

Để \(n^3-n^2+2n+7⋮n^2+1\Leftrightarrow n+8⋮n^2+1\)

\(\Rightarrow\left(n+8\right)\left(n-8\right)⋮n^2+1\Rightarrow n^2-64⋮n^2+1\)

\(\Rightarrow n^2+1-65⋮n^2+1\Rightarrow65⋮n^2+1\)

\(\Rightarrow n^2+1\inƯ\left(65\right)=\left\{\pm1;\pm5;\pm13;\pm65\right\}\)

Mà \(n^2+1\ge1\Rightarrow n^2+1\in\left\{1;5;13;65\right\}\)

\(\Rightarrow n\in\left\{0;\pm2;\sqrt{12};\pm8\right\}\)

Answer 2

[Eren]

Câu c ý tưởng thì hay đó, mỗi tội thiếu bước thử lại