\(1,P=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-1}=\dfrac{3}{2}:\left(-\dfrac{1}{2}\right)=-3\\ 2,A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ b,A=1+\dfrac{1}{\sqrt{x}}>1\left(\dfrac{1}{\sqrt{x}}>0\right)\\ c,\dfrac{P}{A}\left(x-1\right)=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ =\sqrt{x}\left(\sqrt{x}+1\right)=0\Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)
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