\(x-1>0\Rightarrow x>1\)
Xét \(x^2-2mx+1\le0\)
TH1: \(\left\{{}\begin{matrix}\Delta'=m^2-1=0\\-\frac{b}{2a}=m>1\end{matrix}\right.\) \(\Rightarrow\) ko tồn tại m thỏa mãn
TH2: \(\left\{{}\begin{matrix}\Delta'=m^2-1>0\\m-\sqrt{m^2-1}>1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-1>0\\m-1>\sqrt{m^2-1}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m>1\\\left(m-1\right)^2>m^2-1\end{matrix}\right.\)
\(\Rightarrow m>1\)