Đáp án:
$\begin{array}{l}
{x^2} - x + 1 = 2\sqrt {3x - 1} \left( {dk:x \ge \frac{1}{3}} \right)\\
\Leftrightarrow {x^2} + 2x - 3x + 1 = 2\sqrt {3x - 1} \\
\Leftrightarrow {x^2} + 2x + 1 = 3x + 2\sqrt {3x - 1} \\
\Leftrightarrow {\left( {x + 1} \right)^2} = 3x - 1 + 2\sqrt {3x - 1} + 1\\
\Leftrightarrow {\left( {x + 1} \right)^2} = {\left( {\sqrt {3x - 1} + 1} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = \sqrt {3x - 1} + 1\\
x + 1 = - \sqrt {3x - 1} - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \sqrt {3x - 1} \\
x + \sqrt {3x - 1} = - 2\left( {loại} \right)
\end{array} \right.\\
\Leftrightarrow {x^2} = 3x - 1\\
\Leftrightarrow {x^2} - 3x + 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{3 + \sqrt 5 }}{2}\\
x = \frac{{3 - \sqrt 5 }}{2}
\end{array} \right.\left( {tm} \right)
\end{array}$
