Lời giải:
ĐK: \(x\geq 1\)
Đặt \(\sqrt{x-1}=a(a\geq 0)\Rightarrow x=a^2+1\)
PT \(\Leftrightarrow a^2+1-4a=6\)
\(\Leftrightarrow a^2-4a-5=0\)
\(\Leftrightarrow (a+1)(a-5)=0\Rightarrow \left[\begin{matrix} a=-1\\ a=5\end{matrix}\right.\Rightarrow a=5\) (do $a\geq 0$)
\(\Rightarrow x=a^2+1=5^2+1=26\)