Lời giải:
ĐK: $x\geq \frac{1}{3}$
PT $\Leftrightarrow (x^2-3x+1)-(x-\sqrt{3x-1})=0$
$\Leftrightarrow (x^2-3x+1)-\frac{x^2-3x+1}{x+\sqrt{3x-1}}=0$
$\Leftrightarrow (x^2-3x+1)\left(1-\frac{1}{x+\sqrt{3x-1}}\right)=0$
Nếu $x^2-3x+1=0$
$\Rightarrow x=\frac{3\pm \sqrt{5}}{2}$
Nếu $1-\frac{1}{x+\sqrt{3x-1}}=0$
$\Rightarrow x+\sqrt{3x-1}=1$
$\Rightarrow \sqrt{3x-1}=1-x$
\(\Rightarrow \left\{\begin{matrix} 1-x\geq 0\\ 3x-1=(1-x)^2=x^2-2x+1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 1\\ x^2-5x+2=0\end{matrix}\right.\)
\(\Rightarrow x=\frac{5-\sqrt{17}}{2}\)
Vậy..............
