\(B=\dfrac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ B=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\\ c,P=B:A=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\\ P=\dfrac{\sqrt{x}+2-5}{\sqrt{x}+2}=1-\dfrac{5}{\sqrt{x}+2}\)
Ta có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\dfrac{5}{\sqrt{x}+2}\le\dfrac{5}{2}\)
\(\Leftrightarrow P=1-\dfrac{5}{\sqrt{x}+2}\ge1-\dfrac{5}{2}=-\dfrac{3}{2}\\ P_{min}=-\dfrac{3}{2}\Leftrightarrow x=0\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}+2}+\dfrac{9\sqrt{x}-10}{4-x}\)
\(=\dfrac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
\(P=\dfrac{B}{A}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}:\dfrac{\sqrt{x}-2}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}.\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=1-\dfrac{5}{\sqrt{x}+2}\)
Do \(\sqrt{x}+2\ge2\Rightarrow\dfrac{5}{\sqrt{x}+2}\le\dfrac{5}{2}\Rightarrow-\dfrac{5}{\sqrt{x}+2}\ge-\dfrac{5}{2}\)
\(\Rightarrow P=1-\dfrac{5}{\sqrt{x}+2}\ge1-\dfrac{5}{2}=-\dfrac{3}{2}\)
\(minP=-\dfrac{3}{2}\Leftrightarrow x=0\)
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