$P = \bigg(\sqrt x - \dfrac{x+2}{\sqrt x + 1}\bigg) : bigg(\dfrac{\sqrt x}{\sqrt x + 1} - \dfrac{\sqrt x - 4}{1 - x}\bigg) (x \geq 0; x \neq 1 ; x\neq 4)$
$P = \dfrac{\sqrt x(\sqrt x + 1) - x - 2}{\sqrt x + 1} : \bigg(\dfrac{\sqrt x}{\sqrt x + 1} - \dfrac{4 - \sqrt x}{(\sqrt x +1)(\sqrt x - 1)}$
$P = \dfrac{x + \sqrt x - x - 2}{\sqrt x + 1} : \dfrac{\sqrt x(\sqrt x - 1) - 4 + \sqrt x}{(\sqrt x +1)(\sqrt x - 1)}$
$P = \dfrac{\sqrt x - 2}{\sqrt x +1} : \dfrac{x - \sqrt x - 4 + \sqrt x}{(\sqrt x +1)(\sqrt x - 1)}$
$P = \dfrac{\sqrt x - 2}{\sqrt x + 1} . \dfrac{(\sqrt x +1)(\sqrt x -1)}{x - 4}$
$P = (\sqrt x - 2) . \dfrac{\sqrt x - 1}{(\sqrt x + 2)(\sqrt x - 2)}$
$P = \dfrac{\sqrt x - 1}{\sqrt x + 2}$
b) $P = \dfrac{\sqrt x - 1}{\sqrt x + 2} = \dfrac{\sqrt x + 2 - 3}{\sqrt x + 2} = 1 - \dfrac{3}{\sqrt x + 2}$
Vì $\sqrt x \geq 0 ∀ x$
nên $\sqrt x + 2 \geq 2 ∀ x$
⇒ $\dfrac{3}{\sqrt x + 2} \leq \dfrac{3}{2}$
⇒ $1 - \dfrac{3}{\sqrt x + 2} \geq 1 - \dfrac{3}{2}$
hay $P \geq \dfrac{-1}{2}$
Dấu "=" xảy ra ⇔ $\sqrt x = 0 ⇔ x = 0 (T/m)$
Vậy $P_{min} = \dfrac{-1}{2} ⇔ x = 0$
