a: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
b: \(x=\sqrt{2}+1+3-\sqrt{2}+12=16\)
Khi x=16 thì \(A=\dfrac{4}{16-4+1}=\dfrac{4}{13}\)
c. \(A=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{\left(x-\sqrt{x}+1\right)-\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}=1-\dfrac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le1\)
Vậy \(MinA=1\), đạt tại \(\sqrt{x}-1=0\Leftrightarrow x=1\)