To rationalize the denominator of the expression \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{6}}\), we can multiply the numerator and the denominator by \(\sqrt{6}\):
\[ \frac{\sqrt{2} + \sqrt{3}}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{(\sqrt{2} + \sqrt{3})\sqrt{6}}{6} \]
Now we can simplify the numerator:
\[ (\sqrt{2} + \sqrt{3})\sqrt{6} = \sqrt{2} \cdot \sqrt{6} + \sqrt{3} \cdot \sqrt{6} = \sqrt{12} + \sqrt{18} \]
Next, we simplify \(\sqrt{12}\) and \(\sqrt{18}\):
\[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \] \[ \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \]
Now we can substitute these back into our expression:
\[ \sqrt{12} + \sqrt{18} = 2\sqrt{3} + 3\sqrt{2} \]
So, the numerator becomes \(2\sqrt{3} + 3\sqrt{2}\). Therefore, the whole expression is:
\[ \frac{2\sqrt{3} + 3\sqrt{2}}{6} \]
We can separate the terms in the numerator:
\[ \frac{2\sqrt{3}}{6} + \frac{3\sqrt{2}}{6} = \frac{\sqrt{3}}{3} + \frac{\sqrt{2}}{2} \]
Thus, the final simplified expression is:
\[ \frac{\sqrt{3}}{3} + \frac{\sqrt{2}}{2} \]
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