To rationalize the denominator of \(\frac{6}{\sqrt{12}}\), we follow these steps:

1. **Simplify the square root in the denominator**:
   \[
   \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}
   \]
   So the expression becomes:
   \[
   \frac{6}{\sqrt{12}} = \frac{6}{2\sqrt{3}} = \frac{6}{2} \times \frac{1}{\sqrt{3}} = 3 \times \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}}
   \]

2. **Multiply both numerator and denominator by \(\sqrt{3}\) to rationalize the denominator**:
   \[
   \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3}
   \]

3. **Simplify the fraction**:
   \[
   \frac{3\sqrt{3}}{3} = \sqrt{3}
   \]

So, \(\frac{6}{\sqrt{12}} = \sqrt{3}\).

Would you like more details or have any questions? Here are 5 related questions to deepen your understanding:

1. How do you simplify square roots like \(\sqrt{48}\)?
2. Can you rationalize the denominator of \(\frac{5}{\sqrt{8}}\)?
3. What happens when there’s a sum or difference in the denominator, like \(\frac{1}{\sqrt{2} + \sqrt{3}}\)?
4. How would you rationalize a denominator in complex numbers, such as \(\frac{1}{1+i}\)?
5. Can you derive a general formula for rationalizing \(\frac{a}{\sqrt{b}}\)?

**Tip**: Always simplify the square root before rationalizing the denominator—it makes calculations easier!

Learn how to rationalize the denominator of the fraction 6/√12 through a step-by-step solution. This problem covers simplifying square roots and rationalizing denominators, key concepts in algebra suitable for students in Grades 8-10.

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