ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ge0\\\sqrt{x-1}-1\ge0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}x-1\ge0\\x-1\ge1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge1\\x\ge2\end{matrix}\right.\)
=> \(x\ge2\)
Ta có : \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\)
=> \(\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{x-1}-1\)
=> \(\sqrt{\left(\sqrt{x-1}-1\right)^2}=\sqrt{x-1}-1\)
=> \(\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
=> \(\left[{}\begin{matrix}\sqrt{x-1}-1=\sqrt{x-1}-1\\\sqrt{x-1}-1=1-\sqrt{x-1}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}0=0\\2\sqrt{x-1}=2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}0=0\\x-1=1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}0=0\\x=2\end{matrix}\right.\) ( TM )
Vậy phương trình có vô số nghiệm \(x\in R\)
