\(x^2-2mx+1+\left|x-m\right|=0\)
\(\Leftrightarrow\left(x-m\right)^2+\left|x-m\right|=0\)
\(\left\{{}\begin{matrix}\left|x-m\right|=-1\\\left|x-m\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-m=-1\\m-x=-1\\x-m=0\\m-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m-1\\x=m+1\\x=m\\x=m\end{matrix}\right.\)
Vậy phương trình có nghiệm x = m-1;x=m+1; x= m