The common denominator desired is $\sqrt{2}\cdot\sqrt{3} = \sqrt{6}$. So, this expression becomes $\frac{\sqrt{2}\cdot(\sqrt{2}\cdot\sqrt{3})+1\cdot\sqrt{3}+\sqrt{3}\cdot(\sqrt{2}\cdot\sqrt{3})+1\cdot\sqrt{2}}{\sqrt{6}}$. Simplifying this gives $\frac{2\sqrt{3}+\sqrt{3}+3\sqrt{2}+\sqrt{2}}{\sqrt{6}} = \frac{4\sqrt{2}+3\sqrt{3}}{\sqrt{6}}$. To rationalize, multiply numerator and denominator by $\sqrt{6}$ to get $\frac{4\sqrt{2}\sqrt{6}+3\sqrt{3}\sqrt{6}}{6}$. Simplifying yields ${\frac{9\sqrt{2}+8\sqrt{3}}{6}}$, so the desired sum is $9+8+6=\boxed{23}$.
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
If we write $\sqrt{2}+\frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{3}}$ in the form $\dfrac{a\sqrt{2} + b\sqrt{3}}{c}$ such that $a$, $b$, and $c$ are positive integers and $c$ is as small as possible, then what is $a+b+c$?
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
23
If we write $\sqrt{2}+\frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{3}}$ in the form $\dfrac{a\sqrt{2} + b\sqrt{3}}{c}$ such that $a$, $b$, and $c$ are positive integers and $c$ is as small as possible, then what is $a+b+c$?
test