\(x^2-7x+6=\left(6-x\right)\left(1-x\right)\)
Để biểu thức có nghĩa\(\Leftrightarrow\left(6-x\right)\left(1-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6-x\ge0\\1-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}6-x\le0\\1-x\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le1\\x\ge6\end{matrix}\right.\)