ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 952

\[1)\ \left\{ \begin{matrix} 3^{x + y + 1} + 7 \bullet 3^{y - 2} = 8 \\ \sqrt{x + y^{2}} = x + y\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\sqrt{x + y^{2}} = x + y\]

\[x + y^{2} = x^{2} + 2xy + y^{2}\]

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

\[x(x + 2y - 1) = 0\]

\[x_{1} = 0;\ \ \ x_{2} = 1 - 2y.\]

\[3^{x + y + 1} + 7 \bullet 3^{y - 2} = 8\]

\[3^{0 + y + 1} + 7 \bullet 3^{y - 2} = 8\]

\[3 \bullet 3^{y} + 7 \bullet \frac{3^{y}}{3^{2}} = 8\]

\[3^{y}\left( 3 + \frac{7}{9} \right) = 8\]

\[3^{y} \bullet \frac{34}{9} = 8\]

\[3^{y} = \frac{36}{17}\]

\[y = \log_{3}\frac{36}{17}.\]

\[3^{1 - 2y + y + 1} + 7 \bullet 3^{y - 2} = 8\]

\[3^{2} \bullet 3^{- y} + 7 \bullet \frac{3^{y}}{3^{2}} - 8 = 0\]

\[\frac{7}{9} \bullet 3^{y} - 8 + \frac{9}{3^{y}} = 0\]

\[7 \bullet 3^{2y} - 72 \bullet 3^{y} + 81 = 0\]

\[D = 5184 - 2268 = 2916\]

\[3_{1}^{y} = \frac{72 - 54}{2 \bullet 7} = \frac{9}{7};\ \]

\[3_{2}^{y} = \frac{72 + 54}{2 \bullet 7} = 9.\]

\[y_{1} = \log_{3}\frac{9}{7};\ y_{2} = \log_{3}9 = 2.\]

\[x_{1} = 1 - 2\log_{3}\frac{9}{7};\]

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

\[x_{1} = \log_{3}3 - \log_{3}\frac{81}{49} = \log_{3}\frac{49}{27}.\]

\[Область\ определения:\]

\[x + y \geq 0\]

\[x \geq - y.\]

\[Ответ:\ \ \left( 0;\ \log_{3}\frac{36}{17} \right);\ \]

\[\left( \log_{3}\frac{49}{27};\ \log_{3}\frac{9}{7} \right).\]

\[2)\ \left\{ \begin{matrix} 2^{x + y + 1} + 7 \bullet 2^{y - 5} = 4 \\ \sqrt{2x + y^{2}} = x + y\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\sqrt{2x + y^{2}} = x + y\]

\[2x + y^{2} = x^{2} + 2xy + y^{2}\]

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

\[x(x + 2y - 2) = 0\]

\[x_{1} = 0;\ \ \ x_{2} = 2 - 2y.\]

\[1)\ 2^{0 + y + 1} + 7 \bullet 2^{y - 5} = 4\]

\[2 \bullet 2^{y} + 7 \bullet \frac{2^{y}}{2^{5}} = 4\]

\[2^{y}\left( 2 + \frac{7}{32} \right) = 4\]

\[2^{y} \bullet \frac{71}{32} = 4\]

\[2^{y} = \frac{128}{71}\]

\[y = \log_{2}\frac{128}{71}.\]

\[2)\ 2^{2 - 2y + y + 1} + 7 \bullet 2^{y - 5} = 4\]

\[2^{3} \bullet 2^{- y} + 7 \bullet \frac{2^{y}}{2^{5}} - 4 = 0\]

\[\frac{7}{32} \bullet 2^{y} - 4 + \frac{8}{2^{y}} = 0\]

\[7 \bullet 2^{2y} - 128 \bullet 2^{y} + 256 = 0\]

\[D = 16\ 384 - 7168 = 9216\]

\[2_{1}^{y} = \frac{128 - 96}{2 \bullet 7} = \frac{16}{7};\]

\[2_{2}^{y} = \frac{128 + 96}{2 \bullet 7} = 16;\]

\[y_{1} = \log_{2}\frac{16}{7}\text{\ \ }и\ \ y_{2} = \log_{2}16 = 4;\]

\[y_{1} = \log_{2}16 - \log_{2}7 = 4 - \log_{2}7.\]

\[x_{1} = 2 - 8 + 2\log_{2}7 = 2\log_{2}7 - 6;\]

\[x_{2} = 2 - 2 \bullet 4 = 2 - 8 = - 6.\]

\[Область\ определения:\]

\[x + y \geq 0\]

\[x \geq - y.\]

\[Ответ:\ \ \left( 0;\ \log_{2}\frac{128}{71} \right);\ \]

\[\left( 2\log_{2}7 - 6;\ 4 - \log_{2}7 \right).\]