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

\[\boxed{\mathbf{13.}}\]

\[\textbf{а)}\lg\left( x^{2} + 1 \right) + 1 = \cos\text{πx}\]

\[\lg\left( x^{2} + 1 \right) = \cos\text{πx} - 1\]

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

\[\cos\text{πx} - 1 \leq 0.\]

\[\left\{ \begin{matrix} \lg\left( x^{2} + 1 \right) = 0 \\ \cos\text{πx} - 1 = 0 \\ \end{matrix} \right.\ \]

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

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

\[x^{2} = 0\]

\[x = 0.\]

\[Проверим:\]

\[\cos(\pi \cdot 0) - 1 = 0\]

\[1 - 1 = 0\]

\[0 = 0.\]

\[Ответ:x = 0.\]

\[\textbf{б)}\lg\left( 1 + |x - 2| \right) + 2 =\]

\[= \left| 1 + \cos\text{πx} \right|\]

\[\lg\left( 1 + |x - 2| \right) =\]

\[= \left| 1 + \cos\text{πx} \right| - 2\]

\[\lg\left( 1 + |x - 2| \right) \geq 0;\]

\[\left| 1 + \cos\text{πx} \right| - 2 \leq 0.\]

\[\left\{ \begin{matrix} \lg\left( 1 + |x - 2| \right) = 0\ \ \\ \left| 1 + \cos\text{πx} \right| - 2 = 0 \\ \end{matrix} \right.\ \]

\[\lg\left( 1 + |x - 2| \right) = \lg 1\]

\[1 + |x - 2| = 1\]

\[|x - 2| = 0\]

\[x = 2.\]

\[Проверим:\]

\[\left| 1 + \cos{\pi \cdot 2} \right| - 2 = 0\]

\[|1 + 1| - 2 = 0\]

\[2 - 2 = 0\]

\[0 = 0.\]

\[Ответ:x = 2.\]

\[\textbf{в)}\ 3 - \lg\left( x^{2} - 10x + 26 \right) =\]

\[= \sqrt{x^{2} - 10x + 34}\]

\[3 - \lg\left( (x - 5)^{2} + 1 \right) =\]

\[= \sqrt{(x - 5)^{2} + 9}\]

\[\lg\left( (x - 5)^{2} + 1 \right) \geq 0\]

\[3 - \lg\left( (x - 5)^{2} + 1 \right) \leq 3.\]

\[\sqrt{(x - 5)^{2}} \geq 0\]

\[\sqrt{(x - 5)^{2} + 9} \geq 3.\]

\[\left\{ \begin{matrix} 3 - \lg\left( (x - 5)^{2} + 1 \right) = 3 \\ \sqrt{(x - 5)^{2} + 9} = 3\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[- \lg\left( (x - 5)^{2} + 1 \right) = 3 - 3\]

\[\lg\left( (x - 5)^{2} + 1 \right) = 0\]

\[\lg\left( (x - 5)^{2} + 1 \right) = \lg 1\]

\[(x - 5)^{2} + 1 = 1\]

\[(x - 5)^{2} = 0\]

\[x = 5.\]

\[Проверим:\]

\[\sqrt{(5 - 5)^{2} + 9} = \sqrt{9} = 3.\]

\[Ответ:x = 5.\]

\[\textbf{г)}\ 2 - \lg\left( 1 + |x - 6| \right) =\]

\[= \sqrt{x^{2} - 12x + 40}\]

\[2 - \lg\left( 1 + |x - 6| \right) =\]

\[= \sqrt{(x - 6)^{2} + 4}\]

\[\lg\left( 1 + |x - 6| \right) \geq 0\]

\[2 - \lg\left( 1 + |x - 6| \right) \leq 2.\]

\[\sqrt{(x - 6)^{2}} \geq 0\]

\[\sqrt{(x - 6)^{2} + 4} \geq 2\]

\[\left\{ \begin{matrix} 2 - \lg\left( 1 + |x - 6| \right) = 2 \\ \sqrt{(x - 6)^{2} + 4} = 2\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(x - 6)^{2} + 4 = 4\]

\[(x - 6)^{2} = 0\]

\[x = 6.\]

\[Проверим:\]

\[2 - \lg\left( 1 + |6 - 6| \right) = 2 - \lg 1 =\]

\[= 2 - 0 = 2.\]

\[Ответ:x = 6.\]