\(\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\ge1-\frac{3}{2}=-\frac{1}{2}\)
"=" khi x = 0
ĐKXĐ: \(x\ge0\)
Đặt \(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\)\(\ge1-\frac{3}{2}=\frac{-1}{2}\)
Vậy min A = \(\frac{-1}{2}\Leftrightarrow x=0\)(thỏa mãn)