Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1426

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\[1)\ \left\{ \begin{matrix} 2^{x + y} = 32\ \ \\ 3^{3y - x} = 27 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 2^{x + y} = 2^{5}\text{\ \ } \\ 3^{3y - x} = 3^{3} \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} x + y = 5\ \ \\ 3y - x = 3 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 5 - y\ \ \\ x = 3y - 3 \\ \end{matrix} \right.\ \]

\[5 - y = 3y - 3\]

\[- 4y = - 8\]

\[y = 2;\]

\[x = 5 - 2 = 3.\]

\[Ответ:\ \ (3;\ 2).\]

\[2)\ \left\{ \begin{matrix} 3^{x} - 2^{2y} = 77 \\ 3^{\frac{x}{2}} - 2^{y} = 7\ \ \ \ \\ \end{matrix} \right.\ \]

\[a = 3^{\frac{x}{2}};\text{\ \ }b = 2^{y}:\]

\[\left\{ \begin{matrix} a^{2} - b^{2} = 77 \\ a - b = 7\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} a^{2} - b^{2} - 77 = 0 \\ a = 7 + b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(7 + b)^{2} - b^{2} - 77 = 0\]

\[49 + 14b + b^{2} - b^{2} - 77 = 0\]

\[14b - 28 = 0\]

\[b - 2 = 0\]

\[b = 2;\]

\[a = 7 + 2 = 9.\]

\[1)\ 3^{\frac{x}{2}} = 9\]

\[3^{\frac{x}{2}} = 3^{2}\]

\[\frac{x}{2} = 2\]

\[x = 4.\]

\[2)\ 2^{y} = 2\]

\[y = 1.\]

\[Ответ:\ \ (4;\ 1).\]

\[3)\ \left\{ \begin{matrix} 3^{x} \bullet 2^{y} = 576\ \ \ \ \ \ \ \\ \log_{\sqrt{2}}(y - x) = 4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 3^{x} \bullet 2^{y} = 9 \bullet 64\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \log_{\sqrt{2}}(y - x) = \log_{\sqrt{2}}\left( \sqrt{2} \right)^{4} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 3^{x} \bullet 2^{y} = 3^{2} \bullet 2^{6} \\ y - x = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} 3^{x} \bullet 2^{y} = 3^{2} \bullet 2^{6} \\ y = x + 4\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[3^{x} \bullet 2^{x + 4} = 3^{2} \bullet 2^{6}\]

\[3^{x} \bullet 2^{x} = 3^{2} \bullet 3^{2}\]

\[x = 2;\]

\[y = 2 + 4 = 6.\]

\[Ответ:\ \ (2;\ 6).\]

\[4)\ \left\{ \begin{matrix} \lg x + \lg y = 4 \\ x^{\lg y} = 1000\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \lg\text{xy} = \lg 10^{4}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \log_{x}x^{\lg y} = \log_{x}1000 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} xy = 10^{4}\text{\ \ \ \ \ \ \ \ \ \ } \\ \lg y = \log_{x}10^{3} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} y = \frac{10^{4}}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \lg y = \log_{x}10^{3} \\ \end{matrix} \right.\ \]

\[\lg\frac{10^{4}}{x} = \log_{x}10^{3}\]

\[\lg 10^{4} - \lg x = \frac{\lg 10^{3}}{\lg x}\]

\[4 - \lg x = \frac{3}{\lg x}\ \ \ \ \ | \bullet \lg x\]

\[4\lg x - \lg^{2}x = 3\]

\[a = \lg x:\]

\[4a - a^{2} = 3\]

\[a^{2} - 4a + 3 = 0\]

\[D = 16 - 12 = 4\]

\[a_{1} = \frac{4 - 2}{2} = 1;\]

\[a = \frac{4 + 2}{2} = 3.\]

\[1)\ \lg x = 1;\]

\[\lg x = \lg 10^{1}\]

\[x = 10;\]

\[y = \frac{10^{4}}{10} = 10^{3} = 1000.\]

\[2)\ \lg x = 3\]

\[\lg x = \lg 10^{3}\]

\[x = 1000;\]

\[y = \frac{10^{4}}{1000} = 10.\]

\[Ответ:\ \ (10;\ 1000);\ \ (1000;\ 10).\]