a: Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)
\(=\dfrac{x+1}{\sqrt{x}}\)
b: Để A<0 thì x+1<0
hay x<-1(loại

