ĐKXĐ: $x>0$
$\frac{x+\sqrt{x}+1}{\sqrt{x}}=\frac{13}3$
$⇒3x+3\sqrt{x}+3=13\sqrt{x}$
$⇔3x-10\sqrt{x}+3=0$
$⇔3x-\sqrt{x}-9\sqrt{x}+3=0$
$⇔\sqrt{x}(3\sqrt{x}-1)-3(3\sqrt{x}-1)=0$
$⇔(\sqrt{x}-3)(3\sqrt{x}-1)=0$
\(⇔\left[ \begin{array}{l}\sqrt{x}-3=0\\3\sqrt{x}-1=0\end{array} \right.⇔\) \(\left[ \begin{array}{l}x=9(Tmđk)\\x=\frac{1}9(Tmđk)\end{array} \right.\)
Vậy $x=9;\frac{1}9$