ĐK: $x \geq 0$
Ta có
$P = \left( \dfrac{\sqrt{x}}{\sqrt{x} + 1} - \dfrac{1}{x + \sqrt{x}} \right) . \left( \dfrac{1}{\sqrt{x} + 1} + \dfrac{1}{x-1} \right)$
$= \left( \dfrac{\sqrt{x}}{\sqrt{x} + 1} - \dfrac{1}{\sqrt{x}(\sqrt{x} + 1)} \right) . \left( \dfrac{1}{\sqrt{x} + 1} + \dfrac{1}{(\sqrt{x}-1)(\sqrt{x} + 1)} \right)$
$= \dfrac{x - 1}{\sqrt{x}(\sqrt{x} + 1)} . \dfrac{\sqrt{x}-1 + 1}{(\sqrt{x}-1)(\sqrt{x} + 1)}$
$= \dfrac{(x-1)\sqrt{x}}{\sqrt{x}(x-1)(\sqrt{x} + 1)}$
$= \dfrac{1}{\sqrt{x} + 1}$
Vậy ta có
$A = \dfrac{1}{\sqrt{x} + 1} + \dfrac{\sqrt{x} + 19}{9}$
$= \dfrac{1}{\sqrt{x} + 1} + \dfrac{\sqrt{x} + 1}{9} + \dfrac{18}{9}$
$= \dfrac{1}{\sqrt{x} + 1} + \dfrac{\sqrt{x} + 1}{9} + 2$
Áp dụng BĐT Cauchy ta có
$\dfrac{1}{\sqrt{x} + 1} + \dfrac{\sqrt{x} + 1}{9} \geq 2 \sqrt{\dfrac{1}{\sqrt{x} + 1} . \dfrac{\sqrt{x} + 1}{9}} = \dfrac{2}{3}$
$<-> \dfrac{1}{\sqrt{x} + 1} + \dfrac{\sqrt{x} + 1}{9} + 2 \geq \dfrac{8}{3}$
$<-> A \geq \dfrac{8}{3}$
Dấu "=" xảy ra khi
$\dfrac{1}{\sqrt{x} + 1} = \dfrac{\sqrt{x} + 1}{9}$
$<-> (\sqrt{x} + 1)^2 = 9$
$<-> \sqrt{x} + 1 = 3$
$<-> \sqrt{x} = 2$
$<-> x = 4$
Vậy GTNN của A là $\dfrac{8}{3}$ đạt đc khi $x = 4$.
