52. $$\sqrt{18}-\frac{3}{\sqrt{3}-6}-\sqrt{6}(2-\sqrt{3})(1+\sqrt{2})$$

52. Решим пример:

• $$\sqrt{18}-\frac{3}{\sqrt{3}-6}-\sqrt{6}(2-\sqrt{3})(1+\sqrt{2}) = 3\sqrt{2} - \frac{3(\sqrt{3}+6)}{(\sqrt{3}-6)(\sqrt{3}+6)} - \sqrt{6}(2 + 2\sqrt{2} - \sqrt{3} - \sqrt{6}) = 3\sqrt{2} - \frac{3(\sqrt{3}+6)}{3-36} - 2\sqrt{6} - 2\sqrt{12} + 3\sqrt{2} + 6 = 3\sqrt{2} - \frac{3(\sqrt{3}+6)}{-33} - 2\sqrt{6} - 2 \cdot 2\sqrt{3} + 3\sqrt{2} + 6 = 6\sqrt{2} + \frac{\sqrt{3}+6}{11} - 2\sqrt{6} - 4\sqrt{3} + 6 = 6\sqrt{2} + \frac{\sqrt{3}}{11} + \frac{6}{11} - 2\sqrt{6} - 4\sqrt{3} + 6 = 6\sqrt{2} + \frac{\sqrt{3}}{11} + \frac{6}{11} - 2\sqrt{6} - \frac{44\sqrt{3}}{11} + \frac{66}{11} = 6\sqrt{2} -\frac{43\sqrt{3}}{11} - 2\sqrt{6} + \frac{72}{11}$$

Ответ:$$6\sqrt{2} -\frac{43\sqrt{3}}{11} - 2\sqrt{6} + \frac{72}{11}$$