Đáp án:
$\begin{array}{l}
Dkxd:x \ge 2\\
P.\sqrt x + x - 1 = 2\sqrt {3x} + 2\sqrt {x - 2} \\
\Rightarrow \frac{{x + 3}}{{\sqrt x }}.\sqrt x + x - 1 = 2\sqrt {3x} + 2\sqrt {x - 2} \\
\Rightarrow 2x + 2 = 2\left( {\sqrt {3x} + \sqrt {x - 2} } \right)\\
\Rightarrow x + 1 = \sqrt {3x} + \sqrt {x - 2} \\
\Rightarrow x = 3\left( {tm} \right)
\end{array}$
