Lời giải:
ĐKXĐ: \(x\geq \frac{1}{3}\)
Ta có: \(x^2-x+1=2\sqrt{3x-1}\)
\(\Leftrightarrow x^2+2x+1=3x+2\sqrt{3x-1}\)
\(\Leftrightarrow (x+1)^2=(3x-1)+2\sqrt{3x-1}+1\)
\(\Leftrightarrow (x+1)^2=(\sqrt{3x-1}+1)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{3x-1}+1\left(1\right)\\x+1=-\sqrt{3x-1}-1\left(2\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow x=\sqrt{3x-1}\Leftrightarrow x^2=3x-1\) (do $x>0$)
\(\Leftrightarrow x^2-3x+1=0\Leftrightarrow x=\frac{3\pm \sqrt{5}}{2}\) (thỏa mãn)
(2) \(\Leftrightarrow x+\sqrt{3x-1}+2=0\)
Điều trên vô lý do
\(x\geq \frac{1}{3}; \sqrt{3x-1}\geq 0\Rightarrow x+\sqrt{3x-1}+2\geq 2+\frac{1}{3}>0\)
Vậy \(x=\frac{3\pm \sqrt{5}}{2}\)
