Решите систему уравнений: x^2+16y^2=73; xy=-6.

Ответ:

\[\left\{ \begin{matrix} x^{2} + 16y^{2} = 73\ \ \ \ \\ xy = - 6\ \ \ \ \ \ \ \ \ \ \ | \cdot 8 \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x^{2} + 16y^{2} = 73 \\ 8xy = - 48\ \ \ \ \ \ \ \ \ \\ \end{matrix}\ \right.\ ( + )\text{\ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} + 8xy + 16y^{2} = 25 \\ xy = - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} (x + 4y)^{2} = 25 \\ xy = - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1.\ \left\{ \begin{matrix} x + 4y = 5 \\ xy = - 6\ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x = 5 - 4y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (5 - 4y)y + 6 = 0\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 5 - 4y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 4y^{2} + 5y + 6 = 0 \\ \end{matrix} \right.\ \]

\[D = 25 + 96 = 121\]

\[y = \frac{- 5 + 11}{- 8} = - \frac{3}{4},\ \ \ \ \ \ \]

\[y = \frac{- 5 - 11}{- 8} = 2\]

\[\left\{ \begin{matrix} x = - 3 \\ y = 2\ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ \ \ } \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} x = 8\ \ \ \ \ \\ y = - \frac{3}{4} \\ \end{matrix} \right.\ \]

\[2.\ \left\{ \begin{matrix} x + 4y = - 5 \\ xy = - 6\ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x = - 5 - 4y\ \ \ \ \ \ \ \ \ \\ ( - 5 - 4y)y = - 6 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x = - 5 - 4y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 4y^{2} - 5y + 6 = 0 \\ \end{matrix} \right.\ \]

\[D = 25 + 96 = 121\]

\[y = \frac{5 + 11}{- 8} = - 2,\ \ \ \ \ \ \ \ \ \ \]

\[y = \frac{5 - 11}{- 8} = \frac{3}{4}\]

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

\[\left\{ \begin{matrix} x = - 8 \\ y = \frac{3}{4}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[Ответ:( - 3;2);\ \ (8;\ - 0,75);\ \ \ \]

\[(3;\ - 2);\ \ ( - 8;0,75).\]