\(Q=\dfrac{x+3\sqrt{x}-6\sqrt{x}-3\sqrt{x}+9}{x-9}=\dfrac{\left(\sqrt{x}-3\right)^2}{x-9}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
Ta có \(A=P.Q=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=1-\dfrac{4}{\sqrt{x}+2}\ge1-\dfrac{4}{2}=-\dfrac{1}{2}\)
Dấu ''='' xảy ra khi x = 0