Найди значение выражения $\sqrt{300} \cos^2{\frac{19\pi}{12}} - \sqrt{75}$.

Решаем: 1. Упростим выражение $\sqrt{300} \cos^2{\frac{19\pi}{12}} - \sqrt{75}$. 2. Преобразуем $\sqrt{300}$ и $\sqrt{75}$: * $\sqrt{300} = \sqrt{100 \cdot 3} = 10\sqrt{3}$ * $\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$ 3. Преобразуем $\cos^2{\frac{19\pi}{12}}$: * $\frac{19\pi}{12} = \frac{12\pi + 7\pi}{12} = \pi + \frac{7\pi}{12}$ * $\cos{\frac{19\pi}{12}} = \cos{(\pi + \frac{7\pi}{12})} = -\cos{\frac{7\pi}{12}}$ * $\cos^2{\frac{19\pi}{12}} = \cos^2{\frac{7\pi}{12}}$ 4. Найдем $\cos{\frac{7\pi}{12}}$: * $\frac{7\pi}{12} = \frac{3\pi}{12} + \frac{4\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3}$ * $\cos{\frac{7\pi}{12}} = \cos{(\frac{\pi}{4} + \frac{\pi}{3})} = \cos{\frac{\pi}{4}}\cos{\frac{\pi}{3}} - \sin{\frac{\pi}{4}}\sin{\frac{\pi}{3}} = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4}$ 5. Тогда $\cos^2{\frac{7\pi}{12}} = (\frac{\sqrt{2} - \sqrt{6}}{4})^2 = \frac{2 - 2\sqrt{12} + 6}{16} = \frac{8 - 4\sqrt{3}}{16} = \frac{2 - \sqrt{3}}{4}$ 6. Подставим в исходное выражение: * $10\sqrt{3} \cdot \frac{2 - \sqrt{3}}{4} - 5\sqrt{3} = \frac{20\sqrt{3} - 30}{4} - 5\sqrt{3} = 5\sqrt{3} - \frac{15}{2} - 5\sqrt{3} = -\frac{15}{2} = -7.5$ **Ответ: -7.5**