Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1427

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\[1)\ \left\{ \begin{matrix} \log_{4}x - \log_{2}y = 0 \\ x^{2} - 5y^{2} + 4 = 0\ \ \ \\ \end{matrix} \right.\ \]

\[\log_{4}x - \log_{2}y = 0\]

\[\log_{2}x^{\frac{1}{2}} - \log_{2}y = 0\]

\[\log_{2}\frac{\sqrt{x}}{y} = \log_{2}2^{0}\]

\[\frac{\sqrt{x}}{y} = 1\]

\[y = \sqrt{x}.\]

\[x^{2} - 5\left( \sqrt{x} \right)^{2} + 4 = 0\]

\[x^{2} - 5x + 4 = 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{5 - 3}{2} = 1;\]

\[x_{2} = \frac{5 + 3}{2} = 4;\]

\[y_{1} = \sqrt{1} = \pm 1;\]

\[y_{2} = \sqrt{4} = \pm 2.\]

\[Имеет\ смысл\ при:\]

\[x > 0\ \ и\ \ y > 0.\]

\[Ответ:\ \ (1;\ 1);\ \ (4;\ 2).\]

\[2)\ \left\{ \begin{matrix} x^{2} + y^{4} = 16\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \log_{2}x + 2\log_{2}y = 3 \\ \end{matrix} \right.\ \]

\[\log_{2}x + \log_{2}y^{2} = 3\]

\[\log_{2}{xy^{2}} = \log_{2}2^{3}\]

\[xy^{2} = 8\]

\[x = \frac{8}{y^{2}}.\]

\[\left( \frac{8}{y^{2}} \right)^{2} + y^{4} = 16\]

\[\frac{64}{y^{4}} + y^{4} = 16\ \ \ \ \ | \bullet y^{4}\]

\[64 + y^{8} = 16y^{4}\]

\[y^{8} - 16y^{4} + 64 = 0\]

\[\left( y^{4} - 8 \right)^{2} = 0\]

\[y^{4} - 8 = 0\]

\[y^{4} = 8\]

\[y = \pm \sqrt[4]{8};\]

\[x = \frac{8}{\left( \pm \sqrt[4]{8} \right)^{2}} = \frac{8}{\sqrt{8}} = \sqrt{8}.\]

\[Имеет\ смысл\ при:\]

\[x > 0\ \ и\ \ y > 0.\]

\[Ответ:\ \ \left( \sqrt{8};\ \ \sqrt[4]{8} \right).\]