$ĐKXĐ: x>0$
$P=\dfrac{1}{x}+\dfrac{9-x}{x+3\sqrt{x}}$
$P=\dfrac{1}{x}+\dfrac{(3+\sqrt{x})(3-\sqrt{x})}{\sqrt{x}(\sqrt{x}+3)}$
$P=\dfrac{1}{x}+\dfrac{3-\sqrt{x}}{\sqrt{x}}$
$P=\dfrac{1+\sqrt{x}(3-\sqrt{x})}{x}$
$P=\dfrac{1+3\sqrt{x}-x}{x}$
Để $\dfrac{P}{Q}\le\dfrac{1}{2}$
$\Leftrightarrow P:Q\le\dfrac{1}{2}$
$\Leftrightarrow\dfrac{1+3\sqrt{x}-x}{x}:\dfrac{\sqrt{x}+1}{\sqrt{x}}\le\dfrac{1}{2}$
$\Leftrightarrow\dfrac{1+3\sqrt{x}-x}{x}.\dfrac{\sqrt{x}}{\sqrt{x}+1}\le\dfrac{1}{2}$
$\Leftrightarrow\dfrac{1+3\sqrt{x}-x}{\sqrt{x}(\sqrt{x}+1)}\le\dfrac{1}{2}$
$\Leftrightarrow\dfrac{1+3\sqrt{x}-x}{\sqrt{x}(\sqrt{x}+1)}-\dfrac{1}{2}\le0$
$\Leftrightarrow\dfrac{2(1+3\sqrt{x}-x)-\sqrt{x}(\sqrt{x}+1)}{2\sqrt{x}(\sqrt{x}+1)}\le0$
$\Leftrightarrow\dfrac{2+6\sqrt{x}-2x-x-\sqrt{x}}{2\sqrt{x}(\sqrt{x}+1)}\le0$
$\Leftrightarrow\dfrac{-3x+5\sqrt{x}+2}{2\sqrt{x}(\sqrt{x}-1)}\le0$
$\Leftrightarrow-3x+5\sqrt{x}+2\le0($vì $: 2\sqrt{x}(\sqrt{x}+1)>0$ với $\forall x>0)$
$\Leftrightarrow(\sqrt{x}-2)(3\sqrt{x}+1)\le0$
$\Leftrightarrow\left[ \begin{array}{l}\sqrt{x}-2\le0\\3\sqrt{x}+1\le0\end{array} \right.$
$\Leftrightarrow\left[ \begin{array}{l}\sqrt{x}\le2\\3\sqrt {x}\le-1(loại)\end{array} \right.$
$\Leftrightarrow x\le4$
Kết hợp với $ĐKXĐ: x>0$ và $x$ nguyên
Vậy $x\in\{1; 2; 3; 4\}$ thì $\dfrac{P}{Q}\le\dfrac{1}{2}$
