a) B = \(\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
B = \(\left(\dfrac{\left(\sqrt{x}-1\right)+\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
B = \(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\) = \(\dfrac{2}{\sqrt{x}+1}\)
b) ta có : B \(\ge\) \(\dfrac{1}{2}\) \(\Leftrightarrow\) \(\dfrac{2}{\sqrt{x}+1}\) \(\ge\) \(\dfrac{1}{2}\) \(\Leftrightarrow\) \(\dfrac{2}{\sqrt{x}+1}\ge\dfrac{2}{4}\)