$A = x - 3\sqrt x + 2$
$= x - 2.\dfrac{3}{2}.\sqrt x + \dfrac{9}{4} - \dfrac{1}{4}$
$= \left(\sqrt x - \dfrac{3}{2}\right)^2 - \dfrac{1}{4}$
Ta có: $\left(\sqrt x - \dfrac{3}{2}\right)^2 \geq 0, \forall x$
$\Leftrightarrow \left(\sqrt x - \dfrac{3}{2}\right)^2 - \dfrac{1}{4} \geq - \dfrac{1}{4}$
Hay $A \geq - \dfrac{1}{4}$
Dấu = xảy ra $\Leftrightarrow \sqrt x = \dfrac{3}{2} \Leftrightarrow x = \dfrac{9}{4}$
Vậy $\min A = -\dfrac{1}{4} \Leftrightarrow x = \dfrac{9}{4}$
$B = \dfrac{2\sqrt x + 4}{\sqrt x + 3}$
$= \dfrac{2(\sqrt x + 3) - 2}{\sqrt x + 3}$
$= 2 - \dfrac{2}{\sqrt x + 3}$
Ta có:
$\sqrt x \geq 0$
$\Leftrightarrow \sqrt x + 3 \geq 3$
$\Leftrightarrow \dfrac{2}{\sqrt x + 3} \leq \dfrac{2}{3}$
$\Leftrightarrow - \dfrac{2}{\sqrt x + 3} \geq -\dfrac{2}{3}$
$\Leftrightarrow 2 - \dfrac{2}{\sqrt x + 3} \geq 2 - \dfrac{2}{3} = \dfrac{4}{3}$
Dấu = xảy ra $\Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0$
Vậy $\min B = \dfrac{4}{3} \Leftrightarrow x = 0$
