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МатематикаРешебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 20
Задание 20
\[\boxed{\mathbf{20.}}\]
\[\textbf{а)}\ \frac{x^{2}}{1 - \cos x} < \frac{x + 2}{1 - \cos x}\]
\[1 - \cos x \neq 0\]
\[\cos x \neq 1\]
\[x \neq 2\pi n.\]
\[M = ( - \infty;0) \cup (0; + \infty).\]
\[x^{2} < x + 2\]
\[x^{2} - x - 2 < 0\]
\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 2\]
\[x_{1} = 2;\ \ x_{2} = - 1;\]
\[(x + 1)(x - 2) < 0\]
\[x \in ( - 1;2).\]
\[Решение\ неравенства:\]
\[x \in ( - 1;0) \cup (0;2).\]
\[Ответ:\ x \in ( - 1;0) \cup (0;2).\]
\[\textbf{б)}\ \frac{x^{2}}{\cos x - 1} > \frac{- x + 2}{\cos x - 1}\]
\[\cos x - 1 \neq 0\]
\[\cos x \neq 1\]
\[x \neq 2\pi n.\]
\[M = ( - \infty;0) \cup (0; + \infty).\]
\[x^{2} > - x + 2\]
\[x^{2} + x - 2 > 0\]
\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = 2\]
\[(x + 1)(x - 2) > 0\]
\[- 2 < x < 1.\]
\[Решение\ неравенства:\]
\[x \in ( - 2;0) \cup (0;1).\]
\[Ответ:\ x \in ( - 2;0) \cup (0;1).\]
\[\textbf{в)}\ \frac{x^{2}}{1 - \sin x} < \frac{2x + 3}{1 - \sin x}\]
\[1 - \sin x \neq 0\]
\[\sin x \neq 1\]
\[x \neq \frac{\pi}{2} + 2\pi n.\]
\[M = \left( - \infty;\frac{\pi}{2} \right) \cup \left( \frac{\pi}{2}; + \infty \right).\]
\[x^{2} < 2x + 3\]
\[x^{2} - 2x - 3 < 0\]
\[D_{1} = 1 + 3 = 4\]
\[x_{1} = 1 + 2 = 3;\]
\[x_{2} = 1 - 2 = - 1;\]
\[(x + 1)(x - 3) < 0\]
\[- 1 < x < 3.\]
\[Решение\ неравенства:\]
\[x \in \left( - 1;\frac{\pi}{2} \right) \cup \left( \frac{\pi}{2};3 \right).\]
\[Ответ:\ x \in \left( - 1;\frac{\pi}{2} \right) \cup \left( \frac{\pi}{2};3 \right).\]
\[\textbf{г)}\ \frac{x^{2}}{\sin x - 1} > \frac{x + 2}{\sin x - 1}\]
\[\sin x - 1 \neq 0\]
\[\sin x \neq 1\]
\[x \neq \frac{\pi}{2} + 2\pi n.\]
\[M = \left( - \infty;\frac{\pi}{2} \right) \cup \left( \frac{\pi}{2}; + \infty \right).\]
\[x^{2} > x + 2\]
\[x^{2} - x - 2 > 0\]
\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 2\]
\[x_{1} = 2;\ \ x_{2} = - 1.\]
\[- 1 < x < 2.\]
\[Решение\ неравенства:\]
\[x \in \left( - 1;\frac{\pi}{2} \right) \cup \left( \frac{\pi}{2};2 \right).\]
\[Ответ:\ x \in \left( - 1;\frac{\pi}{2} \right) \cup \left( \frac{\pi}{2};2 \right).\]
