\(2\sqrt{x-1}=\sqrt{2x+5}\\ đk\left\{{}\begin{matrix}x-1\ge0\\2x+5\ge0\end{matrix}\right.=>x\ge1\\ \Rightarrow\left(2\sqrt{x-1}\right)^2=\left(\sqrt{2x+5}\right)^2\\ \Rightarrow4\left(x-1\right)=2x+5\\ \Rightarrow4x-4-2x-5=0\\ \Rightarrow2x-9=0\\ \Rightarrow x=\dfrac{9}{2}\left(thoamanđk\right)\)
\(2\sqrt{x-1}=\sqrt{2x+5}\)
ĐK: \(x\ge1;x\ge-\dfrac{5}{2}\Rightarrow x\ge1\)
PT \(\Leftrightarrow\left(2\sqrt{x-1}\right)^2=\left(\sqrt{2x+5}\right)^2\)
\(\Leftrightarrow4\left(x-1\right)=2x+5\)
\(\Leftrightarrow4x-4=2x+5\)
\(\Leftrightarrow4x-2x=5+4\)
\(\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\) (TMĐK)
Vậy \(S=\left\{\dfrac{9}{2}\right\}.\)
