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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 3
Задание 3
\[\boxed{\mathbf{3.}}\]
\[\textbf{а)}\ e^{\sqrt{1 - x^{2}}} + \sqrt{x^{2} - 7x - 8} \geq - 6\]
\[\left\{ \begin{matrix} 1 - x^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \\ x^{2} - 7x - 8 \geq 0 \\ \end{matrix} \right.\ \]
\[1 - x^{2} \geq 0\]
\[x^{2} - 1 \leq 0\]
\[(x + 1)(x - 1) \leq 0\]
\[- 1 \leq x \leq 1.\]
\[x^{2} - 7x - 8 \geq 0\]
\[x_{1} + x_{2} = 7;\ \ x_{1} \cdot x_{2} = - 8\]
\[x_{1} = - 1;\ \ x_{2} = 8;\]
\[(x + 1)(x - 8) \geq 0\]
\[x \leq - 1;\ \ x \geq 8.\]
\[Единственное\ решение\ \]
\[системы:\]
\[x = - 1.\]
\[Проверка:\]
\[e^{0} + 0 \geq - 6\]
\[1 \geq - 6.\]
\[Ответ:x = - 1.\]
\[\textbf{б)}\ \pi^{\sqrt{4 - x^{2}}} + \sqrt{x^{2} - x - 6} \geq 6\]
\[\left\{ \begin{matrix} 4 - x^{2} \geq 0\ \ \ \ \ \ \ \\ x^{2} - x - 6 \geq 0 \\ \end{matrix} \right.\ \]
\[4 - x^{2} \geq 0\]
\[x^{2} - 4 \leq 0\]
\[(x + 2)(x - 2) \leq 0\]
\[- 2 \leq x \leq 2.\]
\[x^{2} - x - 6 \geq 0\]
\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 6\]
\[x_{1} = - 2;\ \ x_{2} = 3;\]
\[(x + 2)(x - 3) \geq 0\]
\[x \leq - 2;\ \ x \geq 3.\]
\[Единственное\ решение:\]
\[x = - 2.\]
\[Проверка:\]
\[\pi^{0} + 0 \geq 6\]
\[1 \geq 6 - неверно.\]
\[Ответ:нет\ корней\]
\[\left\{ \begin{matrix} x^{2} - 4x - 5 \geq 0\ \ \ \ \ \ \ \ \ \\ - 2x^{2} + 8x + 10 \geq 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 4x - 5 \geq 0\]
\[D_{1} = 4 + 5 = 9\]
\[x_{1} = 2 + 3 = 5;\]
\[x_{2} = 2 - 3 = - 1;\]
\[(x + 1)(x - 5) \geq 0\]
\[x \leq - 1;\ \ \ x \geq 5.\]
\[- 2x^{2} + 8x + 10 \geq 0\ \ |\ :( - 2)\]
\[x^{2} - 4x - 5 \leq 0\]
\[(x + 1)(x - 5) \leq 0\]
\[- 1 \leq x \leq 5.\]
\[Решения\ системы:\]
\[x = - 1;\ \ x = 5.\]
\[Проверка.\]
\[x = 5:\]
\[0 + \lg 1 \leq 6\]
\[0 \leq 6\]
\[верно.\]
\[x = - 1:\]
\[0 + \lg 1 \leq 6\]
\[0 \leq 6\]
\[верно.\]
\[Ответ:x = - 1;\ \ x = 5.\]
\[\left\{ \begin{matrix} x^{2} - x - 6 \geq 0 \\ 4 - x^{2} \geq 0\ \ \ \ \ \ \ \ \\ 12 + x > 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} - x - 6 \geq 0\]
\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 6\]
\[x_{1} = - 2;\ \ x_{2} = 3;\]
\[(x + 2)(x - 3) \geq 0\]
\[x \leq - 2;\ \ x \geq 3.\]
\[4 - x^{2} \geq 0\]
\[x^{2} - 4 \leq 0\]
\[(x + 2)(x - 2) \leq 0\]
\[- 2 \leq x \leq 2.\]
\[12 + x > 0\]
\[x > - 12.\]
\[Единственное\ решение:\]
\[x = - 2.\]
\[Проверка:\]
\[0 + 10^{0} + 5\lg 10 \leq 6\]
\[6 \leq 6\]
\[верно.\]
\[Ответ:x = - 2.\]