a, \(P=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\) (ĐK: \(x\ge0,x\ne4,x\ne9\))
\(=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b, ĐK: \(x\ge0,x\ne4,x\ne9\)
\(\dfrac{1}{P}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-4}{\sqrt{x}+1}=1-\dfrac{4}{\sqrt{x}+1}\)
Ta có: \(x\ge0\forall x\in TXĐ\Leftrightarrow\sqrt{x}\ge0\)\(\Leftrightarrow\sqrt{x}+1\ge1\Leftrightarrow\dfrac{1}{\sqrt{x}+1}\le1\Leftrightarrow\dfrac{4}{\sqrt{x}+1}\le4\)\(\Leftrightarrow-\dfrac{4}{\sqrt{x}+1}\ge-4\Leftrightarrow1-\dfrac{4}{\sqrt{x}+1}\ge-3\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
Vậy GTNN của \(\dfrac{1}{P}=-3\Leftrightarrow x=0\)
