Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 29

\[\boxed{\mathbf{29.}}\]

\[\textbf{а)}\ tgx + x^{2} < tgx + 2x + 3\]

\[x \neq \frac{\pi}{2} + \pi k.\]

\[x^{2} < 2x + 3\]

\[x^{2} - 2x - 3 < 0\]

\[D_{1} = 1 + 3 = 4\]

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

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

\[(x + 1)(x - 3) < 0\]

\[- 1 < x < 3.\]

\[Решение\ неравенства:\]

\[x \in \left( - 1;\frac{\pi}{2} \right) \cup \left( \frac{\pi}{2};3 \right).\]

\[Ответ:\ x \in \left( - 1;\frac{\pi}{2} \right) \cup \left( \frac{\pi}{2};3 \right).\]

\[\textbf{б)}\ tgx + x^{2} + 2x > tgx + 3\]

\[x \neq \frac{\pi}{2} + \pi k.\]

\[x^{2} + 2x > 3\]

\[x^{2} + 2x - 3 > 0\]

\[D_{1} = 1 + 3 = 4\]

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

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

\[(x + 3)(x - 1) > 0\]

\[x < - 3;\ \ x > 1.\]

\[Решение\ неравенства:\]

\(Ответ:\ \)

\[\textbf{в)}\ ctg\ x + x^{2} < ctgx + x + 6\]

\[\text{ctg\ x} \neq \text{πk.}\]

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

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

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

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

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

\[- 2 < x < 3.\]

\[Решение\ неравенства:\]

\[x \in ( - 2;0) \cup (0;3).\]

\[Ответ:\ x \in ( - 2;0) \cup (0;3).\]

\[\textbf{г)}\ ctg\ x + x^{2} + x > ctg\ x + 6\]

\[\text{ctg\ x} \neq \text{πk.}\]

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

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

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

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

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

\[x < - 3;\ \ x > 2.\]

\[Решение\ неравенства:\]

\(Ответ:\ \)