đk x >= 0 ; x khác 1
\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\left(\dfrac{2\sqrt{x}+x-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\left(\dfrac{1}{x+\sqrt{x}+1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+2}\)
\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\\ =\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}\right)^3-1}-\dfrac{x+\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\\ =\left(\dfrac{2\sqrt{x}+x}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{x+\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{1}{\sqrt{x}+2}\)
