đkxđ:x>=0
\(A^2=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x+1}\right)^2}=\frac{x-2\sqrt{x}+1}{x+1}=1-\frac{2\sqrt{x}}{x+1}\)
vì \(\left(\sqrt{x}-1\right)^2=x-2\sqrt{x}+1>=0\Rightarrow x+1>=2\sqrt{x}\)
\(\Rightarrow\frac{2\sqrt{x}}{x+1}< =\frac{x+1}{x+1}=1\Rightarrow1-\frac{2\sqrt{x}}{x+1}>=1-1=0\)
dấu = xảy ra khi x=1
vậy min A là 0 khi x-=1