1. (4 балла) Вычислите: 1) 32^(1/5) * 64^(1/2) - 125^(1/3); 2) log_12(1/2) + log_12(1/72); 3) (sqrt[4]{1024}) / (sqrt[3]{1/4}); 4) 5tg(pi/4) - 6cos(pi/3).

1. Вычисление выражений:

1. 1) \( 32^{\frac{1}{5}} \cdot 64^{\frac{1}{2}} - 125^{\frac{1}{3}} \)
   \( = \sqrt[5]{32} \cdot \sqrt{64} - \sqrt[3]{125} \)
   \( = 2 \cdot 8 - 5 \)
   \( = 16 - 5 = 11 \)
2. 2) \( \log_{12}\frac{1}{2} + \log_{12}\frac{1}{72} \)
   Используем свойство логарифмов \( \log_a x + \log_a y = \log_a (x \cdot y) \):
   \( = \log_{12} \left( \frac{1}{2} \cdot \frac{1}{72} \right) \)
   \( = \log_{12} \frac{1}{144} \)
   Так как \( 12^2 = 144 \), то \( \frac{1}{144} = 12^{-2} \).
   \( = \log_{12} 12^{-2} = -2 \)
3. 3) \( \frac{\sqrt[4]{1024}}{\sqrt[3]{\frac{1}{4}}} \)
   \( 1024 = 2^{10} \), \( \sqrt[4]{1024} = \sqrt[4]{2^{10}} = 2^{\frac{10}{4}} = 2^{\frac{5}{2}} \)
   \( \frac{1}{4} = 2^{-2} \), \( \sqrt[3]{\frac{1}{4}} = \sqrt[3]{2^{-2}} = 2^{-\frac{2}{3}} \)
   \( \frac{2^{\frac{5}{2}}}{2^{-\frac{2}{3}}} = 2^{\frac{5}{2} - (-\frac{2}{3})} = 2^{\frac{5}{2} + \frac{2}{3}} = 2^{\frac{15+4}{6}} = 2^{\frac{19}{6}} \)
4. 4) \( 5\operatorname{tg}\frac{\pi}{4} - 6\cos\frac{\pi}{3} \)
   \( \operatorname{tg}\frac{\pi}{4} = 1 \)
   \( \cos\frac{\pi}{3} = \frac{1}{2} \)
   \( = 5 \cdot 1 - 6 \cdot \frac{1}{2} \)
   \( = 5 - 3 = 2 \)

Ответ: 1) 11; 2) -2; 3) 2^19/6; 4) 2.