1. (4 балла) Вычислите: 1) 32^{1/2} \(\cdot\) 64^{1/3} - 125^{1/3}; 2) \(\log\)_{12} \(\frac{1}{2}\) + \(\log\)_{12} \(\frac{1}{72}\);

Решение:

1. \( 32^{1/2} = \sqrt{32} = 4\sqrt{2} \), \( 64^{1/3} = \sqrt[3]{64} = 4 \), \( 125^{1/3} = \sqrt[3]{125} = 5 \).
   \( 4\sqrt{2} \cdot 4 - 5 = 16\sqrt{2} - 5 \)
2. \( \log_{12} \frac{1}{2} + \log_{12} \frac{1}{72} = \log_{12} (\frac{1}{2} \cdot \frac{1}{72}) = \log_{12} \frac{1}{144} = \log_{12} 12^{-2} = -2 \)

Ответ: 1) \( 16\sqrt{2} - 5 \); 2) \( -2 \).