a: \(A=\left(\dfrac{2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}-2\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
=>0<x<4
Vậy: 0<x<4 và x<>1