Выясни, каким числом является значение выражения: a) $(\sqrt{2}+\sqrt{3})-(\sqrt{2}-\sqrt{3})$

- a) $(\sqrt{2} + \sqrt{3}) \cdot (\sqrt{2} - \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$ *Перевод: $(\sqrt{2} + \sqrt{3}) \cdot (\sqrt{2} - \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$* - б) $(\sqrt{2} + 2\sqrt{3}) \cdot (\sqrt{2} - \sqrt{3}) = \sqrt{2} \cdot \sqrt{2} - \sqrt{2} \cdot \sqrt{3} + 2\sqrt{3} \cdot \sqrt{2} - 2\sqrt{3} \cdot \sqrt{3} = 2 + \sqrt{2} \cdot \sqrt{3} - 2 \cdot 3 = 2 + \sqrt{6} - 6 = \sqrt{6} - 4$ *Перевод: $(\sqrt{2} + 2\sqrt{3}) \cdot (\sqrt{2} - \sqrt{3}) = \sqrt{2} \cdot \sqrt{2} - \sqrt{2} \cdot \sqrt{3} + 2\sqrt{3} \cdot \sqrt{2} - 2\sqrt{3} \cdot \sqrt{3} = 2 + \sqrt{2} \cdot \sqrt{3} - 2 \cdot 3 = 2 + \sqrt{6} - 6 = \sqrt{6} - 4$* - г) $\frac{1}{\sqrt{3} - \sqrt{2}} - \frac{1}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2}) - (\sqrt{3} - \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2}}{3 - 2} = \frac{2\sqrt{2}}{1} = 2\sqrt{2}$ *Перевод: $\frac{1}{\sqrt{3} - \sqrt{2}} - \frac{1}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2}) - (\sqrt{3} - \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2}}{3 - 2} = \frac{2\sqrt{2}}{1} = 2\sqrt{2}$* - д) $\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}} = \frac{(\sqrt{2} + \sqrt{3})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{(\sqrt{2} + \sqrt{3})^2}{3 - 2} = \frac{2 + 2\sqrt{2}\sqrt{3} + 3}{1} = 5 + 2\sqrt{6}$ *Перевод: $\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}} = \frac{(\sqrt{2} + \sqrt{3})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{(\sqrt{2} + \sqrt{3})^2}{3 - 2} = \frac{2 + 2\sqrt{2}\sqrt{3} + 3}{1} = 5 + 2\sqrt{6}$*