Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 22

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\[\textbf{а)}\ \frac{2}{\sqrt{\sin x}} > \frac{x^{2} - x}{\sqrt{\sin x}}\]

\[\sin x > 0\]

\[2\pi n < x < \pi + 2\pi n.\]

\[M = (0;\ \pi).\]

\[2 > x^{2} - x\]

\[x^{2} - x - 2 < 0\]

\[x_{1} + x_{2} = 1;x_{1} \cdot x_{2} = - 2\]

\[x_{1} = - 1;\ \ \ x_{2} = 2.\]

\[(x + 1)(x - 2) < 0\]

\[- 1 < x < 2.\]

\[Решение\ неравенства:\]

\[x \in (0;2).\]

\[Ответ:\ x \in (0;2).\]

\[\textbf{б)}\ \frac{6}{\sqrt{\cos x}} > \frac{x^{2} + x}{\sqrt{\cos x}}\]

\[\cos x > 0\]

\[- \frac{\pi}{2} + 2\pi n < x < \frac{\pi}{2} + 2\pi n\ .\]

\[M = \left( - \frac{\pi}{2};\frac{\pi}{2} \right).\]

\[6 > x^{2} + x\]

\[x^{2} + x - 6 < 0\]

\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = - 3;\ \ x_{2} = 2;\]

\[(x + 3)(x - 2) < 0\]

\[- 3 < x < 2.\]

\[Решение\ неравенства:\]

\[x \in \left( - \frac{\pi}{2};\frac{\pi}{2} \right).\]

\[Ответ:\ x \in \left( - \frac{\pi}{2};\frac{\pi}{2} \right).\]