$$ \frac{6}{3-\sqrt{11}} $$ $$ \frac{6}{3-\sqrt{11}}=\square $$ Rationalize the denominator.

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Calculate the value by following steps:
- step0: Calculate:
  $$\frac{6}{3-\sqrt{11}}$$
- step1: Multiply by the Conjugate:
  $$\frac{6\left(3+\sqrt{11}\right)}{\left(3-\sqrt{11}\right)\left(3+\sqrt{11}\right)}$$
- step2: Multiply the numbers:
  $$\frac{6\left(3+\sqrt{11}\right)}{-2}$$
- step3: Factor the expression:
  $$\frac{3\left(3+\sqrt{11}\right)\times 2}{-2}$$
- step4: Reduce the fraction:
  $$\frac{3\left(3+\sqrt{11}\right)}{-1}$$
- step5: Divide the numbers:
  $$-3\left(3+\sqrt{11}\right)$$
- step6: Apply the distributive property:
  $$-3\times 3-3\sqrt{11}$$
- step7: Multiply the numbers:
  $$-9-3\sqrt{11}$$
To rationalize the denominator of the expression $$ \frac{6}{3-\sqrt{11}} $$, we multiply the numerator and the denominator by the conjugate of the denominator, which is $$ 3+\sqrt{11} $$.

Here are the steps:

1. **Multiply by the conjugate:**
   $$
   \frac{6}{3-\sqrt{11}} \cdot \frac{3+\sqrt{11}}{3+\sqrt{11}} = \frac{6(3+\sqrt{11})}{(3-\sqrt{11})(3+\sqrt{11})}
   $$

2. **Calculate the denominator:**
   $$
   (3-\sqrt{11})(3+\sqrt{11}) = 3^2 - (\sqrt{11})^2 = 9 - 11 = -2
   $$

3. **Calculate the numerator:**
   $$
   6(3+\sqrt{11}) = 18 + 6\sqrt{11}
   $$

4. **Combine the results:**
   $$
   \frac{18 + 6\sqrt{11}}{-2} = -9 - 3\sqrt{11}
   $$

Thus, the rationalized form of $$ \frac{6}{3-\sqrt{11}} $$ is:
$$
-9 - 3\sqrt{11}
$$
$$ \frac{6}{3-\sqrt{11}} = -9 - 3\sqrt{11} $$

\( \frac{6}{3-\sqrt{11}} \) \( \frac{6}{3-\sqrt{11}}=\square \) Rationalize the denominator.

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