Giải:
Giả sử \(\left(x-a\right)\left(x-10\right)+1=\) \(\left(x-m\right)\left(x-n\right)\)\(;\left(m,n\in Z\right)\)
\(\Leftrightarrow x^2-\left(a+10\right)x+10a+1\) \(=x^2-\left(m+n\right)x+mn\)
\(\Leftrightarrow\left\{{}\begin{matrix}m+n=a+10\\m.n=10a+1\end{matrix}\right.\) Khử \(a\) ta có:
\(mn=10\left(m+n-10\right)+1\)
\(\Leftrightarrow mn-10m-10n+100=1\)
\(\Leftrightarrow m\left(n-10\right)-10n+100=1\)
Mà \(n,m\in Z\) nên ta có:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}m-10=1\\n-10=1\end{matrix}\right.\\\left\{{}\begin{matrix}m-10=-1\\n-10=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=12\\a=8\end{matrix}\right.\)
Vậy...
