\(\sqrt{x+2\sqrt{x-1}}=2x-1\left(Đk:x\ge1\right)\)
\(\sqrt{x-1+2\sqrt{x-1}+1}=2x-1\)
\(\sqrt{\left(\sqrt{x-1}+1\right)^2}=2x-1\)
\(\sqrt{x-1}+1=2x-1\)
\(\sqrt{x-1}=2\left(x-1\right)\)
\(\sqrt{x-1}\left(2\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\2\sqrt{x-1}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4\left(x-1\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{4}\end{matrix}\right.\)
\(\sqrt{x+2\sqrt{x-1}}=2x-1\) \(\left(x\ge1\right)\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}\right)^2+2\cdot\sqrt{x-1}\cdot1+1^2}=2x-1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2x-1\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2x-1\)
Vì: \(\sqrt{x-1}+1>0.khi.x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+1=2x-1\)
\(\Leftrightarrow\sqrt{x-1}=2x-1-1\)
\(\Leftrightarrow\sqrt{x-1}=2x-2\)
\(\Leftrightarrow\sqrt{x-1}=2\left(x-1\right)\)
\(\Leftrightarrow\sqrt{x-1}-2\left(\sqrt{x-1}\right)^2=0\)
\(\Leftrightarrow\sqrt{x-1}\left(1-2\sqrt{x-1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\1-2\sqrt{x-1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\sqrt{x-1}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{4}+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{4}\end{matrix}\right.\left(tm\right)\)
Vậy: \(S=\left\{1;\dfrac{5}{4}\right\}\)
