Cho biểu thức: 

- Hướng dẫn giải

1. Điều kiện: $0\le x\ne 9$

Ta có:

$\begin{array}{l}P=\frac{2\sqrt{x}(\sqrt{x}-3)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ =\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\frac{-3\sqrt{x}-3}{x-9}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\frac{-3\left(\sqrt{x}+1\right)}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\frac{-3}{\sqrt{x}+3}\end{array}$

2. Ta có:

$\begin{array}{l}P<-\frac{1}{2}\iff \frac{-3}{\sqrt{x}+3}<-\frac{1}{2}\\ \iff \frac{-3}{\sqrt{x}+3}+\frac{1}{2}<0\\ \iff \frac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}<0\\ \iff \sqrt{x}-3<0\\ \iff \sqrt{x}<3\\ \iff 0\le x<9\end{array}$

Vậy $0\le x<9$ thì $P<-\frac{1}{2}$

3. Ta có: $P=\frac{-3}{\sqrt{x}+3}$

Nhận xét

$\begin{array}{l}x\ge 0\iff \sqrt{x}\ge 0\\ \iff \sqrt{x}+3\ge 3\\ \iff \frac{1}{\sqrt{x}+3}\le \frac{1}{3}\\ \iff \frac{-3}{\sqrt{x}+3}\ge -1\\ \iff P\ge -1\end{array}$

Vậy $P_{\min }=-1$ đạt được khi x=0