\(a,P=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{3}{x-2\sqrt{x}}\right):\dfrac{\sqrt{x}-3}{x-4\sqrt{x}+4}\\ =\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)^2}\\ =\left(\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-3}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
\(b,\dfrac{\sqrt{x}-2}{\sqrt{x}}< 0\)
mà \(x>0=>\sqrt{x}>0\)
\(=>\sqrt{x}-2< 0\\ =>\sqrt{x}< 2\\ =>x< 4\)
KH vs đk xđ \(x>0;x\ne4;x\ne9\)
=> \(0< x< 4;x\ne4;x\ne9\)
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