ai làm bài 2 giúp mình với
a) ĐK: $x\ge 0$
$A=\dfrac1{\sqrt x+1}-\dfrac{x+2}{x\sqrt x+1}$
$=\dfrac1{\sqrt x+1}-\dfrac{x+2}{(\sqrt x+1)(x-\sqrt x+1)}$
$=\dfrac{x-\sqrt x+1-x-2}{(\sqrt x+1)(x-\sqrt x+1)}$
$=\dfrac{-(\sqrt x+1)}{(\sqrt x+1)(x-\sqrt x+1)}$
$=\dfrac{-1}{x-\sqrt x+1}$
b) Ta có: $A=\dfrac{-1}{x-\sqrt x+1}=\dfrac{-1}{(\sqrt x-\dfrac12)^2+\dfrac 34}\ge \dfrac{-1}{\dfrac 34}=\dfrac{-4}3$
Dấu '=' xảy ra khi $x=\dfrac14$
Vậy...