ĐXXĐ: \(x\ge1\)
\(\Leftrightarrow\left(x-1\right)^2-6=\left(1-2\left(x-1\right)\right)\sqrt{x-1}\)
Đặt \(\sqrt{x-1}=a\ge0\) ta được:
\(a^4-6=\left(1-2a^2\right)a\Leftrightarrow a^4+2a^3-a-6=0\)
\(\Leftrightarrow\left(a^2+a+2\right)\left(a^2+a-3\right)=0\)
\(\Leftrightarrow a^2+a-3=0\Rightarrow\left[{}\begin{matrix}a=\frac{-1+\sqrt{13}}{2}\\a=\frac{-1-\sqrt{13}}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-1}=\frac{-1+\sqrt{13}}{2}\Rightarrow x=\frac{9-\sqrt{13}}{2}\)