\(x^2-12x+11=0\)
\(\Leftrightarrow x^2-11x-x+11=0\)
\(\Leftrightarrow x\left(x-11\right)-\left(x-11\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
Vậy \(x\in\left\{1;11\right\}\)