\(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2}{x-1}\)
Vậy \(Q=\dfrac{2}{x-1}\)
