\(Đkxđ:x\ge0\)
\(A=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
\(\Rightarrow A=\frac{2\sqrt{x}+2-3}{\sqrt{x+1}}\)
\(\Rightarrow A=\frac{2\left(\sqrt{x}+1\right)-3}{\sqrt{x+1}}\)
\(\Rightarrow A=2-\frac{3}{\sqrt{x}+1}\)
\(\Rightarrow A\ge2-\frac{3}{0+1}\)
\(\Rightarrow A\ge-1\)
Dấu " = " xảy ra \(\Leftrightarrow x=0\)
Vậy: \(Min_A=-1\)