$\dfrac{x}{\sqrt{x} + \sqrt{9 - x}} + \dfrac{x}{\sqrt{x} - \sqrt{9 - x}} = \dfrac{24}{\sqrt{x}}$
$\dfrac{x\left( \sqrt{x} - \sqrt{9 - x} \right) + x\left( \sqrt{x} + \sqrt{9 - x} \right)}{\left( \sqrt{x} + \sqrt{9 - x} \right) \left( \sqrt{x} - \sqrt{9 - x} \right)} = \dfrac{24}{\sqrt{x}}$
$\dfrac{x\sqrt{x} - x\sqrt{9 - x} + x\sqrt{x} + x\sqrt{9 - x}}{x - (9 - x)} = \dfrac{24}{\sqrt{x}}$
$\dfrac{2x\sqrt{x}}{2x - 9} = \dfrac{24}{\sqrt{x}}$
$2x\sqrt{x}\left(\sqrt{x} \right) = 24(2x - 9)$
$x^2 = 12(2x - 9)$
$x^2 - 24x + 108 = 0$
$(x - 18)(x - 6) = 0$
$x = 18 ~ \text{ and/or } 6$
Check the given equation for x = 18
$\dfrac{x}{\sqrt{x} + \sqrt{9 - x}} + \dfrac{x}{\sqrt{x} - \sqrt{9 - x}} = \dfrac{24}{\sqrt{x}}$
$\dfrac{18}{\sqrt{18} + \sqrt{9 - 18}} + \dfrac{18}{\sqrt{18} - \sqrt{9 - 18}} = \dfrac{24}{\sqrt{18}}$
$\dfrac{18}{\sqrt{9(2)} + \sqrt{-9}} + \dfrac{18}{\sqrt{9(2)} - \sqrt{-9}} = \dfrac{3(8)}{\sqrt{9(2)}}$
$\dfrac{18}{3\sqrt{2} + 3i} + \dfrac{18}{3\sqrt{2} - 3i} = \dfrac{3(8)}{3\sqrt{2}}$
$\dfrac{6}{\sqrt{2} + i} + \dfrac{6}{\sqrt{2} - i} = \dfrac{8}{\sqrt{2}}$
$\dfrac{6(\sqrt{2} - i) + 6(\sqrt{2} + i)}{(\sqrt{2} + i)(\sqrt{2} - i)} = \dfrac{8}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}$
$\dfrac{6\sqrt{2} - 6i + 6\sqrt{2} + 6i}{2 - i^2} = \dfrac{8\sqrt{2}}{2}$
$\dfrac{12\sqrt{2}}{2 - (-1)} = 4\sqrt{2}$
$\dfrac{12\sqrt{2}}{3} = 4\sqrt{2}$ ← okay!
Check the given equation for x = 6
$\dfrac{x}{\sqrt{x} + \sqrt{9 - x}} + \dfrac{x}{\sqrt{x} - \sqrt{9 - x}} = \dfrac{24}{\sqrt{x}}$
$\dfrac{6}{\sqrt{6} + \sqrt{9 - 6}} + \dfrac{6}{\sqrt{6} - \sqrt{9 - 6}} = \dfrac{24}{\sqrt{6}}$
$\dfrac{6}{\sqrt{6} + \sqrt{3}} + \dfrac{6}{\sqrt{6} - \sqrt{3}} = \dfrac{24}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}}$
$\dfrac{6(\sqrt{6} - \sqrt{3}) + 6(\sqrt{6} + \sqrt{3})}{(\sqrt{6} + \sqrt{3})(\sqrt{6} - \sqrt{3})} = \dfrac{24\sqrt{6}}{6}$
$\dfrac{6\sqrt{6} - 6\sqrt{3} + 6\sqrt{6} + 6\sqrt{3}}{6 - 3} = 4\sqrt{6}$
$\dfrac{12\sqrt{6}}{3} = 4\sqrt{6}$ ← okay!
Thus, x = 18 and 6 answer
