\(\sqrt{2x-3}=\sqrt{x-1}\left(đk:x\ge\frac{3}{2}\right)\)
\(\Rightarrow\left(\sqrt{2x-3}\right)^2=\left(\sqrt{x-1}\right)^2\)
\(\Leftrightarrow2x-3=x-1\)
\(\Leftrightarrow2x-x=-1+3\)
\(\Leftrightarrow x=2\)(t/m đk )
Vậy Phương trình có nghiệm x=2