Решебник по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 24

\[\boxed{\mathbf{24.}}\]

\[\sqrt{x^{2} + x + 1} + \frac{1}{\sqrt{x^{2} + x + 1}} \geq 2;\]

\[\ln^{2}\left( x^{2} + x + 1 \right) \geq 0\]

\[- \ln^{2}\left( x^{2} + x + 1 \right) \leq 0\]

\[2 - \ln^{2}\left( x^{2} + x + 1 \right) \leq 2.\]

\[\left\{ \begin{matrix} \sqrt{x^{2} + x + 1} + \frac{1}{\sqrt{x^{2} + x + 1}} = 2 \\ 2 - \ln^{2}\left( x^{2} + x + 1 \right) = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sqrt{x^{2} + x + 1} = 1\ \ \ \ \ \ \\ \ln^{2}\left( x^{2} + x + 1 \right) = 0 \\ \end{matrix} \right.\ \]

\[x^{2} + x + 1 = 1\]

\[x^{2} + x = 0\]

\[x(x + 1) = 0\]

\[x = 0;\ \ x = - 1.\]

\[\ln^{2}\left( x^{2} + x + 1 \right) = 0\]

\[x^{2} + x + 1 = 1\]

\[x = 0;\ \ x = - 1.\]

\[Ответ:x = - 1;\ \ x = 0.\]

\[0 \leq \sin^{2}\text{πx} \leq 1\]

\[1 \leq 2 - \sin^{2}\text{πx} \leq 2.\]

\[\left\{ \begin{matrix} \sqrt{x^{2} - 3,5x + 3,5} = 1 \\ \sin^{2}\text{πx} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} - 3,5x + 3,5 = 1\]

\[x^{2} - 3,5x + 2,5 = 0\ \ \ \ | \cdot 2\]

\[2x^{2} - 7x + 5 = 0\]

\[D = 49 - 40 = 9\]

\[x_{1} = \frac{7 + 3}{4} = 2,5;\]

\[x_{2} = \frac{7 - 3}{4} = 1.\]

\[\sin\text{πx} = 0\]

\[\pi x = \pi k\]

\[x = k;\ \ k \in Z.\]

\[Ответ:x = 1.\]