ĐKXĐ : $x\ge \frac{3}{2}$
$\sqrt{2x-3}-\sqrt x= 2x-6$$\Leftrightarrow \frac{2x-3-x}{\sqrt{2x-3}+\sqrt x}=2x-6$
$\Leftrightarrow \frac{x-3}{\sqrt{2x-3}+\sqrt x}=2(x-3)$
$\Leftrightarrow (x-3)(\frac{1}{\sqrt{2x-3}+\sqrt{x}}-2)=0$
Do $x\ge\frac{3}{2} \Rightarrow \sqrt{2x-3}+\sqrt x \ge \sqrt{\frac{3}{2}}\Rightarrow \frac{1}{\sqrt{2x-3}+\sqrt x}\le\sqrt{\frac{2}{3}}$
$\Rightarrow \frac{1}{\sqrt{2x-3}+\sqrt x}-2$ luôn âm
Vậy $x=3$