\(A=\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\)
\(A=\dfrac{1}{\sqrt{x+1}}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}-1}\)
b)
Ta có: A>0
<=>\(\dfrac{1}{\sqrt{x}-1}>0\)
<=>\(\sqrt{x}-1>0\)
<=>\(\sqrt{x}>1\)
<=>\(x>1\)
Vậy x>1 thì A>0