a) \(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(đk:x\ge0,x\ne4,x\ne9\right)\)
\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) \(P=\dfrac{A}{B}=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}-3}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)^2+4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\ge2\sqrt{\left(\sqrt{x}+1\right).\dfrac{4}{\sqrt{x}+1}}=2.2=4\left(Cauchy\right)\)
\(minP=4\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\)
\(\Leftrightarrow\sqrt{x}+1=2\Leftrightarrow x=1\left(tm\right)\)
