Ta có: $\dfrac{\sqrt{x}}{\sqrt{x} + 3} : \left ( \dfrac{9 - x}{x + \sqrt{x} - 6} - \dfrac{\sqrt{x} - 3}{2 - \sqrt{x}} - \dfrac{\sqrt{x} - 2}{\sqrt{x} + 3} \right )$
$= \dfrac{\sqrt{x}}{\sqrt{x} + 3} : \left ( \dfrac{9 - x}{x + \sqrt{x} - 6} + \dfrac{\sqrt{x} - 3}{\sqrt{x} - 2} - \dfrac{\sqrt{x} - 2}{\sqrt{x} + 3} \right )$
$= \dfrac{\sqrt{x}}{\sqrt{x} + 3} : \dfrac{9 - x + \left ( \sqrt{x} - 3 \right )\left ( \sqrt{x} + 3 \right ) - \left ( \sqrt{x} - 2 \right )^{2}}{\left ( \sqrt{x} - 2 \right )\left ( \sqrt{x} + 3 \right )}$
$= \dfrac{\sqrt{x}}{\sqrt{x} + 3} : \dfrac{9 - x + x - 9 - \left ( \sqrt{x} - 2 \right )^{2}}{\left ( \sqrt{x} - 2 \right )\left ( \sqrt{x} + 3 \right )}$
$= \dfrac{\sqrt{x}}{\sqrt{x} + 3} : \left ( -\dfrac{\sqrt{x} - 2}{\sqrt{x} + 3} \right )$
$= -\dfrac{\sqrt{x}}{\sqrt{x} - 2}$
