a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)
Ta có: \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\)
\(\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}\cdot1+1}-\sqrt{x-1}+1=0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}-\sqrt{x-1}+1=0\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|-\sqrt{x-1}+1=0\)
\(\Leftrightarrow\sqrt{x-1}-1-\sqrt{x-1}+1=0\)
\(\Leftrightarrow0x=0\)
Vậy: S={x|\(x\in R\)}