ĐKXĐ: x \(\ge\) 0
(√x - 1):(√x+1) = \(\dfrac{\text{√x}+1}{\text{√x}+1}\) - \(\dfrac{2}{\text{√x}+1}\) = 1 - \(\dfrac{2}{\text{√x}+1}\)
mà x \(\ge\) 0 nên √x \(\ge\)0 => √x+1 \(\ge\)1 => \(\dfrac{2}{\text{√x}+1}\)\(\le\)2 => 1 - \(\dfrac{2}{\text{√x}+1}\)\(\ge\)-1
=> (√x - 1):(√x+1) \(\ge\)-1