Đáp án:
$\begin{array}{l}
Dkxd:x \ge \dfrac{3}{2}\\
\sqrt {2x - 3} - \sqrt x = 2x - 6\\
\Rightarrow \dfrac{{2x - 3 - x}}{{\sqrt {2x - 3} + \sqrt x }} = 2\left( {x - 3} \right)\\
\Rightarrow \dfrac{{x - 3}}{{\sqrt {2x - 3} + \sqrt x }} = 2\left( {x - 3} \right)\\
\Rightarrow \left[ \begin{array}{l}
x - 3 = 0\\
\dfrac{1}{{\sqrt {2x - 3} + \sqrt x }} = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {tmdk} \right)\\
\sqrt {2x - 3} + \sqrt x = \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
2x - 3 + 2\sqrt {x\left( {2x - 3} \right)} + x = \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow 3x - 3 + 2\sqrt {2{x^2} - 3x} = \dfrac{1}{4}\\
\Rightarrow 2\sqrt {2{x^2} - 3x} = \dfrac{{13}}{4} - 3x\left( 1 \right)\\
\left( {dkxd:x \le \dfrac{{13}}{{12}}} \right)\\
Do:Dkxd:x \ge \dfrac{3}{2}\\
\Rightarrow \left( 1 \right)\,vô\,nghiệm
\end{array}$
Vậy x=3
