a) Ta có : \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
=> \(P:Q=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{-3}{\sqrt{x}+3}\)
Để P:Q < -1/2
<=> \(-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{2}\)
<=> \(\dfrac{3}{\sqrt{x}+3}>\dfrac{1}{2}\Leftrightarrow6>\sqrt{x}+3\Leftrightarrow x< 9\)
Kết hợp ĐKXĐ : => Khi \(0\le x< 9\) thì P : Q < -1/2
b) Ta có \(\sqrt{x}+3\ge3\Rightarrow\dfrac{3}{\sqrt{x}+3}\le1\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\ge-1\)
=> Min P:Q = -1
"=" khi x = 0
1: \(P:Q=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
Để P:Q<-1/2 thì \(\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow-6+\sqrt{x}+3< 0\)
hay 0<x<9
2: \(\sqrt{x}+3>=3\forall x\)
\(\Leftrightarrow-\dfrac{3}{\sqrt{x}+3}>=-1\forall x\)
Dấu '=' xảy ra khi x=0
