điều kiện xác đinh : \(x\ge1\)
đặc : \(\sqrt{x-1}=t\left(t\ge0\right)\)
ta có : \(x^2-x+2\sqrt{x-1}=0\Leftrightarrow x\left(x-1\right)+2\sqrt{x-1}=0\)
\(\Leftrightarrow\left(t^2+1\right)t^2+2t=0\Leftrightarrow t^4+t^2+2t=0\Leftrightarrow t\left(t+1\right)\left(t^2-t+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t+1=0\\t^2-t+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}t=0\left(N\right)\\t=-1\left(L\right)\\t\in\varnothing\end{matrix}\right.\)
với \(t=0\Leftrightarrow\sqrt{x-1}=0\Leftrightarrow x=1\) vậy \(x=1\)
