To simplify and rationalize the denominator of the expression \(\frac{\sqrt{3}}{-2 - \sqrt{3}}\), we can multiply the numerator and the denominator by the conjugate of the denominator.
The conjugate of \(-2 - \sqrt{3}\) is \(-2 + \sqrt{3}\).
So, we multiply the numerator and denominator by \(-2 + \sqrt{3}\):
\[ \frac{\sqrt{3}}{-2 - \sqrt{3}} \cdot \frac{-2 + \sqrt{3}}{-2 + \sqrt{3}} = \frac{\sqrt{3}(-2 + \sqrt{3})}{(-2 - \sqrt{3})(-2 + \sqrt{3})} \]
Now, let's calculate the denominator:
\[ (-2 - \sqrt{3})(-2 + \sqrt{3}) = (-2)^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]
Next, calculate the numerator:
\[ \sqrt{3}(-2 + \sqrt{3}) = -2\sqrt{3} + 3 \]
Putting it all together, we have:
\[ \frac{\sqrt{3}}{-2 - \sqrt{3}} = \frac{-2\sqrt{3} + 3}{1} = -2\sqrt{3} + 3 \]
Therefore, the simplified expression is:
\[ 3 - 2\sqrt{3} \]
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