\(a,A=\left(\dfrac{\sqrt{x}-9}{x-3\sqrt{x}}+\dfrac{8}{\sqrt{x}-3}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-3}\right)\left(ĐKXĐ:x>0;x\ne9\right)\)
\(=\left[\dfrac{\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)}+\dfrac{8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]:\left[\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(=\dfrac{9\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{x-9-x}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{9\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-9}\)
\(=-\left(\sqrt{x}-1\right)\)
\(=1-\sqrt{x}\)
\(b,\) Khi đó: \(A=\dfrac{1}{2}\Leftrightarrow1-\sqrt{x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
\(c,\) Ta có: \(\dfrac{1}{A}>\dfrac{2}{3}\)
\(\Leftrightarrow A< \dfrac{3}{2}\)
\(\Leftrightarrow1-\sqrt{x}< \dfrac{3}{2}\)
\(\Leftrightarrow\sqrt{x}>-\dfrac{1}{2}\)
\(\Leftrightarrow x>\dfrac{1}{4}\)
Kết hợp với điều kiện, ta được: \(x>\dfrac{1}{4};x\ne9\)
#\(Toru\)