ĐKXĐ: $x\neq9$
a) $P=\Bigg(\dfrac{2\sqrt[]{x}}{\sqrt[]{x}+3}+\dfrac{\sqrt[]{x}}{\sqrt[]{x}+3}-\dfrac{3x+3}{x-9}\Bigg):\Bigg(\dfrac{2\sqrt[]{x}-2}{\sqrt[]{x}-3}-1\Bigg)$
$=\dfrac{2\sqrt[]{x}(\sqrt[]{x}-3)+\sqrt[]{x}(\sqrt[]{x}-3)-3x-3}{(\sqrt[]{x}+3)(\sqrt[]{x}-3)}:\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}-3}$
$=\dfrac{-9\sqrt[]{x}-3}{(\sqrt[]{x}+3)(\sqrt[]{x}-3)}.\dfrac{\sqrt[]{x}-3}{\sqrt[]{x}+1}$
$=\dfrac{-3(\sqrt[]{x}+1)(3\sqrt[]{x}+1)}{(\sqrt[]{x}+3)(\sqrt[]{x}+1)}$
$=\dfrac{-9\sqrt[]{x}-3}{\sqrt[]{x}+3}$
b) $P<-\dfrac{1}{2}$
$↔ \dfrac{-9\sqrt[]{x}-3}{\sqrt[]{x}+3}<-\dfrac{1}{2}$
$↔ 17\sqrt[]{x}+3>0$
$↔ \sqrt[]{x}>-\dfrac{3}{17}$ (Đúng $∀x≥0$)
c) $P=\dfrac{-9(\sqrt[]{x}+3)}{\sqrt[]{x}+3}+\dfrac{24}{\sqrt[]{x}+3}$
$=-9+\dfrac{24}{\sqrt[]{x}+3}$
$P_{max}$ khi $(\sqrt[]{x}+3)$ nhỏ nhất
Ta có: $\sqrt[]{x}≥0 → \sqrt[]{x}+3≥3$ (Dấu $"="$ xảy ra khi và chỉ khi $x=0$)
$→ P_{max}=-9+\dfrac{24}{3}=-1$
(Không có $P_{min}$)
