a: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{\sqrt{x}+1-\sqrt{x}}{x-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{1}{\sqrt{x}-1}\)
b: Để A<-1/2 thì A+1/2<0
\(\Leftrightarrow\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{2}< 0\)
=>\(\dfrac{2+\sqrt{x}-1}{2\left(\sqrt{x}-1\right)}< 0\)
=>0<=x<1