a) ĐKXĐ: \(x>0;x\ne1\)
b) \(A=\left(\dfrac{1}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)
\(A=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{1}{\sqrt{x}+2}\right]:\dfrac{1-\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\)
\(A=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right]:\dfrac{1-\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\)
\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}\)
\(A=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
c) \(A=\dfrac{5}{3}\) khi và chỉ khi:
\(\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{5}{3}\)
\(\Leftrightarrow3\sqrt{x}+6=5\sqrt{x}\)
\(\Leftrightarrow5\sqrt{x}-3\sqrt{x}=6\)
\(\Leftrightarrow2\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow x=9\left(tm\right)\)
