\(\dfrac{1}{3\sqrt{2}+3\sqrt{4}+1}=\dfrac{1}{7+3\sqrt{2}}=\dfrac{7-3\sqrt{2}}{49-18}=\dfrac{7-3\sqrt{2}}{31}\)
\(\dfrac{1}{3\sqrt{2}+3\sqrt{4}+1}=\dfrac{1}{3\sqrt{2}+3.2+1}=\dfrac{1}{3\sqrt{2}+7}=\dfrac{3\sqrt{2}-7}{\left(3\sqrt{2}+7\right)\left(3\sqrt{2}-7\right)}=\dfrac{3\sqrt{2}-7}{18-49}=\dfrac{7-3\sqrt{2}}{31}\)