\(\dfrac{x+\sqrt{3x}+3}{x\sqrt{x}-3\sqrt{3}}=\dfrac{x+\sqrt{3x}+3}{\left(\sqrt{x}-\sqrt{3}\right)\left(x+\sqrt{3x}+3\right)}=\dfrac{1}{\sqrt{x}-\sqrt{3}}\)
\(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)