a) [tex]A=(\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}}) : (\frac{\sqrt{x}-1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+1})=\frac{2}{x-2\sqrt{x}}:\frac{(\sqrt{x}-1)(\sqrt{x}+1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{2}{\sqrt{x}(\sqrt{x}-2)}:\frac{x-1-(x-4)}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{2}{\sqrt{x}(\sqrt{x}-2)}.\frac{(\sqrt{x}-2)(\sqrt{x}+1)}{3}=\frac{2(\sqrt{x}+1)}{3}[/tex]
b) [tex]B=\frac{2(\sqrt{x}+1)}{3}.\sqrt{x}-x=\frac{2}{3}x+\frac{2}{3}\sqrt{x}-x=\frac{2}{3}\sqrt{x}-\frac{1}{3}x=-\frac{1}{3}(x-2\sqrt{x})=\frac{1}{3}-\frac{1}{3}(x-2\sqrt{x}+1)=\frac{1}{3}-\frac{1}{3}(\sqrt{x}-1)^2\leq \frac{1}{3}[/tex]
Dấu "=" xảy ra tại [tex]x=1[/tex]