4. Вычислите: a) (3√2-2)(4√2+7)-13√2 б) (3√2+2)²+(6-√2)² в) (√11-2√2)(2√2+√11)+(-2√5)2

Ответ:

1. a) $$(3\sqrt{2} - 2)(4\sqrt{2} + 7) - 13\sqrt{2} = 12 \cdot 2 + 21\sqrt{2} - 8\sqrt{2} - 14 - 13\sqrt{2} = 24 + 13\sqrt{2} - 14 - 13\sqrt{2} = 10$$
2. б) $$(3\sqrt{2} + 2)^2 + (6 - \sqrt{2})^2 = (9 \cdot 2 + 12\sqrt{2} + 4) + (36 - 12\sqrt{2} + 2) = 18 + 12\sqrt{2} + 4 + 36 - 12\sqrt{2} + 2 = 60$$
3. в) $$(\sqrt{11} - 2\sqrt{2})(2\sqrt{2} + \sqrt{11}) + (-2\sqrt{5})^2 = (\sqrt{11} - 2\sqrt{2})(\sqrt{11} + 2\sqrt{2}) + 4 \cdot 5 = 11 - 4 \cdot 2 + 20 = 11 - 8 + 20 = 23$$