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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 4
Задание 4
\[\boxed{\mathbf{4.}}\]
\[\left\{ \begin{matrix} x^{2} - 81 \geq 0 \\ 81 - x^{2} \geq 0 \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} - 81 \geq 0\]
\[(x + 9)(x - 9) \geq 0\]
\[x \leq - 9;\ \ x \geq 9.\]
\[81 - x^{2} \geq 0\]
\[x^{2} - 81 \leq 0\]
\[(x + 9)(x - 9) \leq 0\]
\[- 9 \leq x \leq 9.\]
\[Решения\ системы:\]
\[x = - 9;\ \ \ x = 9.\]
\[Проверка.\]
\[x = - 9:\]
\[(0 + 2)\log_{3}| - 9| - (0 + 1) > 4\]
\[3 > 4\]
\[неверно.\]
\[x = 9:\]
\[(0 + 2)\log_{3}|9| - (0 + 1) > 4\]
\[5 > 4\]
\[верно.\]
\[Ответ:x = 9.\]
\[\left\{ \begin{matrix} x^{2} - 16 \geq 0 \\ 16 - x^{2} \geq 0 \\ x^{2} - 7 > 0\ \\ \end{matrix} \right.\ \]
\[x^{2} - 16 \geq 0\]
\[x \leq - 4;\ \ x \geq 4.\]
\[16 - x^{2} \geq 0\]
\[x^{2} - 16 \leq 0\]
\[- 4 \leq x \leq 4.\]
\[x^{2} - 7 > 0\]
\[x^{2} > 7\]
\[x < - \sqrt{7};\ \ x > \sqrt{7}.\]
\[Решения\ системы:\]
\[x = - 4;\ \ x = 4.\]
\[Проверка.\]
\[x = - 4:\]
\[(0 + 2)\log_{3}9 + 2(0 + 3) < 4\]
\[8 < 0\]
\[неверно.\]
\[x = 4:\]
\[(0 + 2)\log_{3}9 - 2(0 + 3) < 4\]
\[- 4 < 0\]
\[верно.\]
\[Ответ:x = 4.\]
\[\left\{ \begin{matrix} x^{2} - 7x + 10 \geq 0 \\ 7x - x^{2} - 10 > 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 7x + 10 \geq 0\]
\[x_{1} + x_{2} = 7;\ \ x_{1} \cdot x_{2} = 10\]
\[x_{1} = 2;\ \ \ x_{2} = 5;\]
\[(x - 2)(x - 5) \geq 0\]
\[x \leq 2;\ \ \ x \geq 5.\]
\[- x^{2} + 7x - 10 > 0\]
\[x^{2} - 7x + 10 < 0\]
\[2 < x < 5.\]
\[Решений\ нет.\]
\[Ответ:нет\ корней.\]
\[\left\{ \begin{matrix} x^{2} + 7x + 10 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - x^{2} - 5x - 6 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 5x + \sqrt{- x^{2} - 5x - 6} > 0 \\ \end{matrix} \right.\ \]
\[x^{2} + 7x + 10 \geq 0\]
\[x_{1} + x_{2} = - 7;\ \ \ x_{1} \cdot x_{2} = 10\]
\[x_{1} = - 5;\ \ \ x_{2} = - 2;\]
\[(x + 5)(x - 2) \geq 0\]
\[x \leq - 5;\ \ \ x \geq - 2.\]
\[- x^{2} - 5x - 6 \geq 0\]
\[x^{2} + 5x + 6 \leq 0\]
\[x_{1} + x_{2} = - 5;\ \ x_{1} \cdot x_{2} = 6\]
\[x_{1} = - 3;\ \ x_{2} = - 2;\]
\[(x + 3)(x + 2) \leq 0\]
\[- 3 \leq x \leq 2.\]
\[Система\ примет\ вид:\]
\[\left\{ \begin{matrix} x = - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 5x + \sqrt{- x^{2} - 5x - 6} > 0 \\ \end{matrix} \right.\ \]
\[Подставим\ x = - 2\ \]
\[в\ неравенство:\]
\[10 + 0 > 0\]
\[10 > 0 - верно.\]
\[Проверка:\]
\[\sqrt{- 4 + 10 - 6} > - 10\]
\[0 > - 10\]
\[верно.\]
\[Ответ:x = - 2.\]