Question

Tìm GTNN của A=\(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)

Answer 1

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