Найдите решение системы уравнений 4*√(x/y)+2*√(y/x)=9; 7√x+2√y=48.

Ответ:

\[\left\{ \begin{matrix} 4\sqrt{\frac{x}{y}} + 2\sqrt{\frac{y}{x}} = 9\ \ \ \\ 7\sqrt{x} + 2\sqrt{y} = 48 \\ \end{matrix} \right.\ \]

\[Пусть\ t = \sqrt{\frac{x}{y}}:\]

\[4t + \frac{2}{t} = 9\]

\[4t^{2} - 9t + 2 = 0\]

\[D = 81 - 32 = 49\]

\[t_{1} = \frac{9 + 7}{8} = 2;\ \ t_{2} = \frac{9 - 7}{8} = \frac{1}{4}.\]

\[1)\ \left\{ \begin{matrix} \sqrt{\frac{x}{y}} = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 7\sqrt{x} + 2\sqrt{y} = 48 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} \sqrt{x} = 2\sqrt{y}\text{\ \ \ \ \ \ \ \ \ \ } \\ 7\sqrt{x} + 2\sqrt{y} = 48 \\ \end{matrix} \right.\ \]

\[7 \cdot 2\sqrt{y} + 2\sqrt{y} = 48\]

\[14\sqrt{y} + 2\sqrt{y} = 48\]

\[16\sqrt{y} = 48\]

\[\sqrt{y} = 3 \Longrightarrow y = 9.\]

\[\sqrt{x} = 2 \cdot 3 = 6 \Longrightarrow x = 36.\]

\[2)\ \left\{ \begin{matrix} \sqrt{\frac{x}{y}} = \frac{1}{4}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 7\sqrt{x} + 2\sqrt{y} = 48 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \sqrt{x} = \frac{1}{4}\sqrt{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ 7\sqrt{x} + 2\sqrt{y} = 48 \\ \end{matrix} \right.\ \]

\[7 \cdot \frac{1}{4}\sqrt{y} + 2\sqrt{y} = 48\]

\[\frac{7}{4}\sqrt{y} + 2\sqrt{y} = 48\]

\[\frac{15}{4}\sqrt{y} = 48\]

\[\sqrt{y} = \frac{12}{15} = \frac{4}{5}\ \Longrightarrow y = \frac{16}{25}.\]

\[\sqrt{x} = \frac{1}{4} \cdot \frac{4}{5} = \frac{1}{5} \Longrightarrow x = \frac{1}{25}.\]

\[Ответ:(36;9);\left( \frac{1}{25};\frac{16}{25} \right).\]