\(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)
\(\dfrac{A}{B}< -\dfrac{1}{2}\Rightarrow\dfrac{-3}{\sqrt{x}+3}< -\dfrac{1}{2}\)
\(\Rightarrow\dfrac{3}{\sqrt{x}+3}>\dfrac{1}{2}\Rightarrow\sqrt{x}+3< 6\)
\(\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)
Kết hợp điều kiện đề bài \(\Rightarrow0\le x< 9\)
