7.1 Найдите значение выражения: $(\sqrt{32} + \sqrt{128}) \cdot \sqrt{2}$.

7.1 $(\sqrt{32} + \sqrt{128}) \cdot \sqrt{2} = (4\sqrt{2} + 8\sqrt{2}) \cdot \sqrt{2} = 12\sqrt{2} \cdot \sqrt{2} = 24$ 7.2 $\frac{(25\sqrt{11})^2}{125} = \frac{625 \cdot 11}{125} = 5 \cdot 11 = 55$ 7.3 $7^{65} : (7^{21})^3 = 7^{65} : 7^{63} = 7^2 = 49$ 7.4 $10^{5,5} : 5^{4,5} \cdot 2^{-3,5} = (2 \cdot 5)^{5,5} : 5^{4,5} \cdot 2^{-3,5} = 2^{5,5} \cdot 5^{5,5} : 5^{4,5} \cdot 2^{-3,5} = 2^{5,5-3,5} \cdot 5^{5,5-4,5} = 2^2 \cdot 5^1 = 4 \cdot 5 = 20$ 7.5 $\frac{\sqrt[3]{21}}{\sqrt[3]{21} \cdot \sqrt[3]{21}} = \frac{1}{\sqrt[3]{21}} = 21^{-1/3}$ 7.6 $\log_9 18,25 + \log_9 4 = \log_9(18,25 \cdot 4) = \log_9 73$ 7.7 $\log_{0,5} 0,7 - \log_{0,5} 1,4 = \log_{0,5}(\frac{0,7}{1,4}) = \log_{0,5} 0,5 = 1$ 7.8 $\frac{\log_{15} 13,5}{\log_{15} 3} + \log_3 2 = \log_3 13,5 + \log_3 2 = \log_3(13,5 \cdot 2) = \log_3 27 = 3$ 7.9 $11^{\log_{\sqrt{17}} 17} = 11^{\log_{17^{0,5}} 17^1} = 11^{2 \cdot \log_{17} 17} = 11^2 = 121$ 7.10 $5\sqrt{2} \cdot \cos(\frac{5\pi}{6}) \cdot \sin(\frac{\pi}{4}) \cdot \operatorname{tg}(-\frac{\pi}{3}) = 5\sqrt{2} \cdot (-\frac{\sqrt{3}}{2}) \cdot \frac{\sqrt{2}}{2} \cdot (-\sqrt{3}) = 5 \cdot \sqrt{2} \cdot \sqrt{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = 5 \cdot 2 \cdot \frac{3}{4} = 10 \cdot 0,75 = 7,5$ 7.11 $7 \cdot \cos(\frac{17\pi}{12}) \cdot \sin(\frac{17\pi}{12}) = \frac{7}{2} \cdot 2 \cdot \cos(\frac{17\pi}{12}) \cdot \sin(\frac{17\pi}{12}) = \frac{7}{2} \cdot \sin(\frac{17\pi}{6}) = 3,5 \cdot \sin(2\pi + \frac{5\pi}{6}) = 3,5 \cdot \sin(\frac{5\pi}{6}) = 3,5 \cdot 0,5 = 1,75$ 7.12 $22\sqrt{3} \cdot \cos^2(\frac{13\pi}{12}) - 11\sqrt{3} = 11\sqrt{3} \cdot (2 \cdot \cos^2(\frac{13\pi}{12}) - 1) = 11\sqrt{3} \cdot \cos(\frac{13\pi}{6}) = 11\sqrt{3} \cdot \cos(2\pi + \frac{\pi}{6}) = 11\sqrt{3} \cdot \cos(\frac{\pi}{6}) = 11\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 11 \cdot 1,5 = 16,5$ 7.13 $\sqrt{12} - \sqrt{48} \cdot \sin^2(\frac{7\pi}{12}) = 2\sqrt{3} - 4\sqrt{3} \cdot \sin^2(\frac{7\pi}{12}) = 2\sqrt{3} \cdot (1 - 2\sin^2(\frac{7\pi}{12})) = 2\sqrt{3} \cdot \cos(\frac{14\pi}{12}) = 2\sqrt{3} \cdot \cos(\frac{7\pi}{6}) = 2\sqrt{3} \cdot (-\frac{\sqrt{3}}{2}) = -3$ 7.14 $17\sqrt{2} \cdot \cos^2(\frac{5\pi}{8}) - 17\sqrt{2} \sin^2(\frac{5\pi}{8}) = 17\sqrt{2} \cdot (\cos^2(\frac{5\pi}{8}) - \sin^2(\frac{5\pi}{8})) = 17\sqrt{2} \cdot \cos(\frac{5\pi}{4}) = 17\sqrt{2} \cdot (-\frac{\sqrt{2}}{2}) = -17$ 7.15 $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - (\frac{3\sqrt{5}}{7})^2} = \sqrt{1 - \frac{9 \cdot 5}{49}} = \sqrt{1 - \frac{45}{49}} = \sqrt{\frac{4}{49}} = \frac{2}{7}$. Так как $\alpha \in (0; \frac{\pi}{2})$, то $\cos \alpha > 0$. Значит, $7 \cdot \cos \alpha = 7 \cdot \frac{2}{7} = 2$ 7.16 $\frac{3 \sin 62^\circ}{\cos 31^\circ \cos 59^\circ} = \frac{3 \sin(2 \cdot 31^\circ)}{\cos 31^\circ \sin(90^\circ - 59^\circ)} = \frac{3 \cdot 2 \sin 31^\circ \cos 31^\circ}{\cos 31^\circ \sin 31^\circ} = 6$