a) \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\left(đk:x\ge0,x\ne9\right)\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
b) \(A=\dfrac{3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{4}+3}=\dfrac{3}{2+3}=\dfrac{3}{5}\)
c) \(A=\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}+3=9\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\left(tm\right)\)
d) \(A=\dfrac{3}{\sqrt{x}+3}>\dfrac{2}{3}\)
\(\Leftrightarrow2\sqrt{x}+6< 9\Leftrightarrow2\sqrt{x}< 3\)
\(\Leftrightarrow\sqrt{x}< \dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{-9}{4}< x< \dfrac{9}{4}\)
Kết hợp với điều kiện:
\(\Rightarrow0\le x< \dfrac{9}{4}\)
