ĐK: \(x\ge0,x\ne4\)
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-\left(5\sqrt{x}+2\right)}{x-4}=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{x-4}=\dfrac{-x+2\sqrt{x}}{x-4}=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}\)\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}:\dfrac{-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)^2}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}.\dfrac{\left(\sqrt{x}+2\right)^2}{-\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
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