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 \(n^2+1\ge1\Rightarrow n^2+1\in\left\{1;5;13;65\right\}\)
\(\Rightarrow n\in\left\{0;\pm2;\sqrt{12};\pm8\right\}\)
\(n^4-2n^3+2n^2-2n+1=\left(n^2-n\right)^2+\left(n-1\right)^2=\left(n-1\right)^2\left(n^2+1\right)\)
Để đ/t trên chia hết cho \(n^4+1\Leftrightarrow\left(n-1\right)^2\left(n^2+1\right)⋮n^4+1\Leftrightarrow\left(n-1\right)^2⋮n^2-1\)
\(\Leftrightarrow\left(n-1\right)\left(n+1-2\right)⋮n^2-1\)
\(\Leftrightarrow n^2-1-2\left(n-1\right)⋮n^2-1\)
\(\Leftrightarrow2\left(n-1\right)⋮n^2-1\)
\(\Leftrightarrow2⋮n+1\)
\(\Leftrightarrow n+1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow n\in\left\{0;-2;1;-3\right\}\)
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