93. Сократите дробь: д) 2√3+3√2-√6 2+√6-√2

д) $$\frac{2\sqrt{3} + 3\sqrt{2} - \sqrt{6}}{2 + \sqrt{6} - \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2})(\sqrt{2} + \sqrt{3} - \sqrt{2} \cdot \sqrt{3})}{2 + \sqrt{6} - \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2})(2\sqrt{3} + 3\sqrt{2} - \sqrt{6})}{2 + \sqrt{6} - \sqrt{2}} = \sqrt{3} + \sqrt{2}$$

Ответ: $$\sqrt{3} + \sqrt{2}$$