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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 16
Задание 16
\[\boxed{\mathbf{16.}}\]
\[\textbf{а)}\ \left| \lg(x - 3) \right| + 2 = \left| \cos\text{πx} + 1 \right|\]
\[\left| \lg(x - 3) \right| = \left| \cos\text{πx} + 1 \right| - 2\]
\[\left| \lg(x - 3) \right| \geq 0;\]
\[- 1 \leq \cos\text{πx} \leq 1\]
\[0 \leq \cos\text{πx} + 1 \leq 2\]
\[0 \leq \left| \cos\text{πx} + 1 \right| \leq 2\]
\[- 2 \leq \left| \cos\text{πx} + 1 \right| - 2 \leq 0.\]
\[\left\{ \begin{matrix} \lg{(x - 3)} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left| \cos\text{πx} + 1 \right| - 2 = 0 \\ \end{matrix} \right.\ \]
\[\lg(x - 3) = 0\]
\[x - 3 = 1\]
\[x = 4.\]
\[Проверим:\]
\[\left| \cos{4\pi} + 1 \right| - 2 = |1 + 1| - 2 =\]
\[= 2 - 2 = 0.\]
\[Ответ:x = 4.\]
\[\textbf{б)}\ \left| \lg(x - 2) \right| + 1 = - \cos\text{πx}\]
\[\left| \lg(x - 2) \right| = - \cos\text{πx} - 1\]
\[\left| \lg(x - 2) \right| \geq 0;\]
\[- 1 \leq \cos\text{πx} \leq 1\]
\[- 1 \leq - \cos\text{πx} \leq 1\]
\[- 2 \leq - \cos\text{πx} - 1 \leq 0.\]
\[\left\{ \begin{matrix} \lg{(x - 2)} = 0\ \ \ \ \ \ \\ - \cos\text{πx} - 1 = 0 \\ \end{matrix} \right.\ \]
\[\lg(x - 2) = 0\]
\[x - 2 = 1\]
\[x = 3.\]
\[Проверим:\]
\[- \cos{3\pi} - 1 = - ( - 1) - 1 =\]
\[= 1 - 1 = 0.\]
\[Ответ:x = 3.\]
\[\textbf{в)}\ \left| \lg(x - 5) \right| + 2 =\]
\[= \sqrt{4 - (x - 6)^{2}}\]
\[\left| \lg(x - 5) \right| = \sqrt{4 - (x - 6)^{2}} - 2\]
\[\left| \lg(x - 5) \right| \geq 0;\]
\[0 \leq \sqrt{4 - (x - 6)^{2}} \leq \sqrt{4}\]
\[- 2 \leq \sqrt{4 - (x - 6)^{2}} - 2 \leq 0.\]
\[\left\{ \begin{matrix} \lg{(x - 5)} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sqrt{4 - (x - 6)^{2}} - 2 = 0 \\ \end{matrix} \right.\ \]
\[\lg(x - 5) = 1\]
\[x - 5 = 1\]
\[x = 6.\]
\[Проверим:\]
\[\sqrt{4 - (6 - 6)^{2}} - 2 = \sqrt{4} - 2 = 0.\]
\[Ответ:x = 6.\ \]
\[\textbf{г)}\ \left| \lg(x - 4) \right| + 3 =\]
\[= \sqrt{9 - (x - 5)^{2}}\]
\[\left| \lg(x - 4) \right| = \sqrt{9 - (x - 5)^{2}} - 3\]
\[\left| \lg(x - 4) \right| \geq 0;\]
\[0 \leq \sqrt{9 - (x - 5)^{2}} \leq \sqrt{9}\]
\[- 3 \leq \sqrt{9 - (x - 5)^{2}} - 3 \leq 0.\]
\[\left\{ \begin{matrix} \left| \lg(x - 4) \right| = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sqrt{9 - (x - 5)^{2}} - 3 = 0 \\ \end{matrix} \right.\ \]
\[\lg(x - 4) = 0\]
\[x - 4 = 1\]
\[x = 5.\]
\[Проверим:\]
\[\sqrt{9 - (5 - 5)^{2}} - 3 = \sqrt{9} - 3 =\]
\[= 3 - 3 = 0.\]
\[Ответ:x = 5.\]