1: Thay x=25 vào B, ta được:
\(B=\dfrac{1}{5-2}=\dfrac{1}{3}\)
2; P=A:B
\(=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{1}\)
\(=\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}\)
\(P=A:B=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\right):\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{x+2-2\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-2\right)\\ =\dfrac{x+2-2x+4\sqrt{x}+x-1}{\sqrt{x}+1}\\ =\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}\)
\(P=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\right):\dfrac{1}{\sqrt{x}-2}\\ =\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right).\left(\sqrt{x}-2\right)\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2x-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}+\dfrac{x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right).\left(\sqrt{x}-2\right)\)
\(=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-2\right)\)
\(=\dfrac{\left(4\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}\)
