Bài 2
b) x+√x +1 =x +√x +$\frac{1}{4}$ +$\frac{3}{4}$ =(√x+$\frac{1}{2}$ )²+$\frac{3}{4}$ >0∀x
=>$\frac{2}{x+Vx +1}$ >0
=>B>0 ∀ x$\neq$ 1
d)
B=$\frac{2}{x+Vx +1}$ =$\frac{2}{(Vx+1/2)^2}$ +$\frac{3}{4}$
(√x+$\frac{1}{2}$ )²+$\frac{3}{4}$ ≥$\frac{3}{4}$
<=>$\frac{2}{(Vx +1/2)^2}$ +$\frac{3}{4}$ ≤$\frac{8}{3}$ <=>B≤$\frac{8}{3}$
→Max B=$\frac{8}{3}$