\(\sqrt{-x^2+4x-3}=2x-5\left(1\le x\le3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{5}{2}\\-x^2+4x-3=4x^2-20x+25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{2}\le x\le3\\5x^2-24x+28=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{14}{5}\left(tmđk\right)\\x=2\left(ktmđk\right)\end{matrix}\right.\)
Vậy \(x=\frac{14}{5}\)