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МатематикаВопрос:
45. $$\frac{-2+\sqrt{5}}{3\sqrt{45}-2\sqrt{5}}-\frac{2-\sqrt{5}}{7}$$
Ответ:
45. Решим пример:
- $$\frac{-2+\sqrt{5}}{3\sqrt{45}-2\sqrt{5}}-\frac{2-\sqrt{5}}{7} = \frac{-2+\sqrt{5}}{3 \cdot 3\sqrt{5}-2\sqrt{5}}-\frac{2-\sqrt{5}}{7} = \frac{-2+\sqrt{5}}{9\sqrt{5}-2\sqrt{5}}-\frac{2-\sqrt{5}}{7} =\frac{-2+\sqrt{5}}{7\sqrt{5}}-\frac{2-\sqrt{5}}{7} = \frac{(-2+\sqrt{5})\cdot 7 - (2-\sqrt{5})\cdot 7\sqrt{5}}{49\sqrt{5}} = \frac{-14 + 7\sqrt{5} - 14\sqrt{5} + 35}{49\sqrt{5}} = \frac{21 - 7\sqrt{5}}{49\sqrt{5}} = \frac{7(3 - \sqrt{5})}{49\sqrt{5}} = \frac{3-\sqrt{5}}{7\sqrt{5}} = \frac{(3-\sqrt{5})\sqrt{5}}{7\sqrt{5}\sqrt{5}} = \frac{3\sqrt{5}-5}{35}$$
Ответ:$$\frac{3\sqrt{5}-5}{35}$$
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