1. (4 балла) Вычислите: 1) 125^{1/3} \(\cdot\) 16^{1/4} - 9^{1/2}; 2) \(\log\)_6 18 - \(\log\)_6 3; 3) \(\frac{\sqrt[3]{256}}{\sqrt{2}}\); 4) 4\(\sin\) \(\frac{\pi}{6}\) + 2\(\cos\) 2\(\pi\).

Решение:

1. \(125^{1/3} \cdot 16^{1/4} - 9^{1/2} = \sqrt[3]{125} \cdot \sqrt[4]{16} - \sqrt{9} = 5 \cdot 2 - 3 = 10 - 3 = 7\)
2. \(\log_6 18 - \log_6 3 = \log_6 \frac{18}{3} = \log_6 6 = 1\)
3. \(\frac{\sqrt[3]{256}}{\sqrt{2}} = \frac{\sqrt[3]{4^3 \cdot 4}}{\sqrt{2}} = \frac{4\sqrt[3]{4}}{\sqrt{2}} = \frac{4 \cdot 2^{2/3}}{2^{1/2}} = 4 \cdot 2^{2/3 - 1/2} = 4 \cdot 2^{1/6} = 2^2 \cdot 2^{1/6} = 2^{13/6}\)
4. \(4\sin \frac{\pi}{6} + 2\cos 2\pi = 4 \cdot \frac{1}{2} + 2 \cdot 1 = 2 + 2 = 4\)

Ответ: 1) 7; 2) 1; 3) \(2^{13/6}\); 4) 4.