A = \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\) ; Đk \(x\) \(\ge\) 0
A = 2 - \(\dfrac{3}{\sqrt{x}+1}\)
\(\sqrt{x}\) \(\ge\) 0 => \(\sqrt{x}\) + 1 \(\ge\) 1
\(\dfrac{3}{\sqrt{x}+1}\) \(\le\) \(\dfrac{3}{1}\)
\(\dfrac{3}{\sqrt{x}+1}\) \(\le\) 3
- \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) -3
2 - \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) 2 - 3
2 - \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) -1
A(min ) = -1 dấu bằng xảy ra khi \(\sqrt{x}\) = 0 hay \(x\) = 0
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