Giải thích các bước giải:

ĐKXĐ: $x\ge 0, x\ne 4$

Ta có:

$P(6x+18)\ge x+9$

$\to P\cdot 6(x+3)\ge x+9$

$\to \dfrac{\sqrt{x}(\sqrt{x}-3)}{(x+3)(\sqrt{x}-2)}\cdot 6(x+3)\ge x+9$

$\to \dfrac{6\sqrt{x}(\sqrt{x}-3)}{\sqrt{x}-2}\ge x+9$

$\to 6x-18\sqrt{x}\ge (\sqrt{x}-2)(x+9)$

$\to 6x-18\sqrt{x}\ge x\sqrt{x}+9\sqrt{x}-2x-18$

$\to x\sqrt{x}-8x+27\sqrt{x}-18\le0$

 $\to \sqrt{x}\le 0.86393$

$\to 0\le x\le 0.86393^2$