ĐK: $x\ge 0$
$P=\dfrac{1-\sqrt{x}}{1+\sqrt{x}+(\sqrt{x})^2}$
$=\dfrac{(1-\sqrt{x})^2}{(1+\sqrt{x}+\sqrt{x}^2)(1-\sqrt{x})}$
$=\dfrac{ (1-\sqrt{x})^2}{1-\sqrt{x}^3}$
Ta có $(1-\sqrt{x})^2\ge 0$
$P<0\to 1-\sqrt{x}^3<0$
$\to \sqrt{x}^3>1$
$\to x>1$
Vậy $x>1$