ГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 46

\[\boxed{\mathbf{46.}}\]

\[\textbf{а)}\ (4 - x)^{x^{2} - 9} - sin^{2}10{^\circ} <\]

\[< (4 - x)^{\frac{1}{\log_{\cos{10{^\circ}}}\sqrt{4 - x}}}\]

\[4 - x > 0\]

\[x < 4.\]

\[4 - x \neq 1\]

\[x \neq 3.\]

\[M = ( - \infty;3) \cup (3;4).\]

\[(4 - x)^{x^{2} - 9} - sin^{2}10{^\circ} < cos^{2}10{^\circ}\ \]

\[(4 - x)^{x^{2} - 9} < 1\]

\[10^{\left( x^{2} - 9 \right)\lg(4 - x)} < 10{^\circ}\ \]

\[\left( x^{2} - 9 \right)\lg(4 - x) < 0\]

\[x < 3:\]

\[x^{2} - 9 < 0\]

\[- 3 < x < 3.\]

\[3 < x < 4:\]

\[x^{2} - 9 > 0\]

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

\[Решение\ неравенства:\]

\[x \in ( - 3;3) \cup (3;4).\]

\[Ответ:x \in ( - 3;3) \cup (3;4).\]

\[\textbf{б)}\ (5 - x)^{x^{2} - 4} - cos^{2}2002{^\circ} <\]

\[< (5 - x)^{\frac{1}{\log_{\sin{2002{^\circ}}}\sqrt{5 - x}}}\]

\[5 - x > 0\]

\[x < 5.\]

\[5 - x \neq 1\]

\[x \neq 4.\]

\[M = ( - \infty;4) \cup (4;5).\]

\[(5 - x)^{x^{2} - 4} - cos^{2}2002{^\circ} <\]

\[< \text{si}n^{2}2002{^\circ}\]

\[(5 - x)^{x^{2} - 4} < 1\]

\[x < 4:\]

\[x^{2} - 4 < 0\]

\[- 2 < x < 2.\]

\[4 < x < 5:\]

\[x^{2} - 4 > 0\]

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

\[Решение\ неравенства:\]

\[x \in ( - 2;2) \cup (4;5).\]

\[Ответ:x \in ( - 2;2) \cup (4;5).\]