Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 30

\[\boxed{\mathbf{30.}}\]

\[\textbf{а)}\ \sqrt{2^{x} - 4} + \sqrt{2^{x} - 8} =\]

\[= \sqrt{2^{x + 1} - 12}\]

\[1)\ \left\{ \begin{matrix} 2^{x} - 4 = 0 \\ 2^{x} - 8 \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2^{x} = 2^{2} \\ 2^{x} \geq 2^{3} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 2 \\ x \geq 3 \\ \end{matrix} \right.\ \]

\[нет\ решений.\]

\[2)\ \left\{ \begin{matrix} 2^{x} - 8 = 0 \\ 2^{x} - 4 \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2^{x} = 2^{3} \\ 2^{x} \geq 2^{2} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 3 \\ x \geq 2 \\ \end{matrix} \right.\ \]

\[Ответ:x = 3.\]

\[\textbf{б)}\ \sqrt{3^{x} - 9} + \sqrt{3^{x} - 3} =\]

\[= \sqrt{2 \cdot 3^{x} - 12}\]

\[1)\ \left\{ \begin{matrix} 3^{x} - 9 = 0 \\ 3^{x} - 3 \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 3^{x} = 3^{2} \\ 3^{x} \geq 3^{1} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 2 \\ x \geq 1 \\ \end{matrix} \right.\ \]

\[x = 2.\]

\[2)\ \left\{ \begin{matrix} 3^{x} - 3 = 0 \\ 3^{x} - 9 \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 3^{x} = 3^{1} \\ 3^{x} \geq 3^{2} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 1 \\ x \geq 2 \\ \end{matrix} \right.\ \]

\[нет\ решений.\]

\[Ответ:x = 2.\]

\[\textbf{в)}\ \sqrt{\log_{5}x - 1} + \sqrt{2^{x} - 2} =\]

\[= \sqrt{\log_{5}x + 2^{x} - 3}\]

\[1)\ \left\{ \begin{matrix} \log_{5}x - 1 = 0 \\ 2^{x} - 2 \geq 0\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \log_{5}x = 1 \\ 2^{x} \geq 2^{1}\text{\ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 5^{1} \\ x \geq 1\ \ \\ \end{matrix} \right.\ \]

\[x = 5.\]

\[2)\ \left\{ \begin{matrix} 2^{x} - 2 = 0\ \ \ \ \ \ \\ \log_{5}x - 1 \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2^{x} = 2^{1}\ \\ x \geq 5^{1}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 1 \\ x \geq 5 \\ \end{matrix} \right.\ \]

\[нет\ решений.\]

\[Ответ:x = 5.\]

\[\textbf{г)}\ \sqrt{\log_{6}x - 1} + \sqrt{3^{x} - 9} =\]

\[= \sqrt{\log_{6}x + 3^{x} - 10}\]

\[1)\ \left\{ \begin{matrix} \log_{6}x - 1 = 0 \\ 3^{x} - 9 \geq 0\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \log_{6}x = 1 \\ 3^{x} \geq 3^{2}\text{\ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 6^{1} \\ x \geq 2\ \ \\ \end{matrix} \right.\ \]

\[x = 6.\]

\[2)\ \left\{ \begin{matrix} 3^{x} - 9 = 0\ \ \ \ \ \\ \log_{6}x - 1 \geq 0 \\ \end{matrix}\ \right.\ \]

\[\left\{ \begin{matrix} 3^{x} = 3^{2}\text{\ \ \ \ } \\ \log_{6}x \geq 1 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 2\ \ \\ x \geq 6^{1} \\ \end{matrix} \right.\ \]

\[нет\ решений.\]

\[Ответ:x = 6.\]