35. Вычислите: 1) \(\sqrt{10} \cdot \sqrt{160}\) 2) \(\sqrt{\frac{1}{25}} \cdot \sqrt{5}\) 3) \(\sqrt{3} \cdot \sqrt{11} \cdot \sqrt{33}\) 4) \((2\sqrt{12} + 2\sqrt{3})^2\) 5) \((2\sqrt{2} - 3\sqrt{3})^2\)

Решение:

1. \(\sqrt{10} \cdot \sqrt{160} = \sqrt{10 \cdot 160} = \sqrt{1600} = 40\)
2. \(\sqrt{\frac{1}{25}} \cdot \sqrt{5} = \frac{1}{5} \cdot \sqrt{5} = \frac{\sqrt{5}}{5}\)
3. \(\sqrt{3} \cdot \sqrt{11} \cdot \sqrt{33} = \sqrt{3 \cdot 11 \cdot 33} = \sqrt{3 \cdot 11 \cdot (3 \cdot 11)} = \sqrt{3^2 \cdot 11^2} = 3 \cdot 11 = 33\)
4. \((2\sqrt{12} + 2\sqrt{3})^2 = (2\sqrt{4 \cdot 3} + 2\sqrt{3})^2 = (2 \cdot 2\sqrt{3} + 2\sqrt{3})^2 = (4\sqrt{3} + 2\sqrt{3})^2 = (6\sqrt{3})^2 = 6^2 \cdot (\sqrt{3})^2 = 36 \cdot 3 = 108\)
5. \((2\sqrt{2} - 3\sqrt{3})^2 = (2\sqrt{2})^2 - 2 \cdot (2\sqrt{2}) \cdot (3\sqrt{3}) + (3\sqrt{3})^2 = (4 \cdot 2) - 12\sqrt{6} + (9 \cdot 3) = 8 - 12\sqrt{6} + 27 = 35 - 12\sqrt{6}\)

Ответ: 1) 40; 2) \(\frac{\sqrt{5}}{5}\); 3) 33; 4) 108; 5) \(35 - 12\sqrt{6}\).