ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 953

\[z = \log_{3}(x + 2y);t = \log_{3}(x - y):\]

\[2z^{2} = zt + t^{2}\]

\[t^{2} + zt - 2z^{2} = 0\]

\[D = z^{2} + 8z^{2} = 9z^{2}\]

\[t_{1} = \frac{- z - 3z}{2} = - 2z;\]

\[t_{2} = \frac{- z + 3z}{2} = z.\]

\[1)\ \log_{3}(x - y) = - 2\log_{3}(x + 2y)\]

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

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

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

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

\[9(x + 2y) = 1\]

\[x + 2y = \frac{1}{9}\]

\[x = \frac{1}{9} - 2y.\]

\[\left( \frac{1}{9} - 2y \right)^{2} + y\left( \frac{1}{9} - 2y \right) - 2y^{2} = 9\]

\[\frac{1}{81} - \frac{4}{9}y + 4y^{2} + \frac{1}{9}y - 2y^{2} - 2y^{2} = 9\]

\[- \frac{1}{3}y = \frac{728}{81}\]

\[y = - \frac{728}{27};\]

\[x = \frac{1}{9} + \frac{1456}{27} = \frac{1459}{27}.\]

\[2)\ \log_{3}(x - y) = \log_{3}(x + 2y)\]

\[x - y = x + 2y\]

\[0 = 3y\]

\[y = 0\]

\[x^{2} + x \bullet 0 - 2 \bullet 0^{2} = 9\]

\[x^{2} = 9\]

\[x = \pm 3.\]

\[Область\ определения:\]

\[x > y;\ \ x > - 2y.\]

\[Ответ:\ \ (3;\ 0);\ \left( \frac{1459}{27};\ - \frac{728}{27} \right).\]

\[z = \log_{2}(x + y);t = \log_{2}(x - 2y):\]

\[z^{2} + zt = 2t^{2}\]

\[z^{2} + zt - 2t^{2} = 0\]

\[D = t^{2} + 4 \bullet 2t^{2} = t^{2} + 8t^{2} = 9t^{2}\]

\[z_{1} = \frac{- t - 3t}{2} = - 2t;\]

\[\ z_{2} = \frac{- t + 3t}{2} = t.\]

\[1)\ \log_{2}(x + y) = - 2\log_{2}(x - 2y)\]

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

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

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

\[\left( x^{2} - xy - 2y^{2} \right)(x - 2y) = 1\]

\[4(x - 2y) = 1\]

\[x - 2y = \frac{1}{4}\]

\[x = \frac{1}{4} + 2y.\]

\[\left( \frac{1}{4} + 2y \right)^{2} - y\left( \frac{1}{4} + 2y \right) - 2y^{2} = 4\]

\[\frac{1}{16} + y + 4y^{2} - \frac{1}{4}y - 2y^{2} - 2y^{2} = 4\]

\[\frac{3}{4}y = \frac{63}{16}\]

\[y = \frac{21}{4};\]

\[x = \frac{1}{4} + \frac{21}{2} = \frac{43}{4}.\]

\[2)\ \log_{2}(x + y) = \log_{2}(x - 2y)\]

\[x + y = x - 2y\]

\[0 = 3y\]

\[y = 0.\]

\[x^{2} - x \bullet 0 - 2 \bullet 0^{2} = 4\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[Область\ определения:\]

\[x > - y;\ \ \ \ x > 2y.\]

\[Ответ:\ \ (2;\ 0);\ \left( \frac{43}{4};\ \frac{21}{4} \right).\]