\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\\left(2x-5-x+1\right)\left(2x-5+x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\\left(x-4\right)\left(3x-6\right)=0\end{matrix}\right.\Leftrightarrow x=4\)
|x-1|=2x-5
=> x-1=2x-5 hoặc x-1=5-2x
-x=-4 3x=6
x=4 x=2 (loại)