17. \( \frac{2 \sqrt{3}-3}{6-\sqrt{3}} \)

To simplify \( \frac{2 \sqrt{3}-3}{6-\sqrt{3}} \), first, we can multiply the numerator and the denominator by the conjugate of the denominator to eliminate the square root. The conjugate of \( 6 - \sqrt{3} \) is \( 6 + \sqrt{3} \). So, we have: \[ \frac{(2\sqrt{3} - 3)(6 + \sqrt{3})}{(6 - \sqrt{3})(6 + \sqrt{3})} \] Calculating the denominator: \[ (6 - \sqrt{3})(6 + \sqrt{3}) = 6^2 - (\sqrt{3})^2 = 36 - 3 = 33 \] Now for the numerator: \[ (2\sqrt{3} - 3)(6 + \sqrt{3}) = 2\sqrt{3} \cdot 6 + 2\sqrt{3} \cdot \sqrt{3} - 3 \cdot 6 - 3 \cdot \sqrt{3} \] \[ = 12\sqrt{3} + 6 - 18 - 3\sqrt{3} \] \[ = (12\sqrt{3} - 3\sqrt{3}) + (6 - 18) = 9\sqrt{3} - 12 \] Now we can combine everything: \[ \frac{9\sqrt{3} - 12}{33} \] We can simplify that further: \[ \frac{3(3\sqrt{3} - 4)}{33} = \frac{3\sqrt{3} - 4}{11} \] So, the simplified form of \( \frac{2 \sqrt{3}-3}{6-\sqrt{3}} \) is: \[ \frac{3\sqrt{3} - 4}{11} \]