Решите систему уравнений: x^2+xy-12y^2=0; 2x^2-3xy+y^2=90.

Ответ:

\[\left\{ \begin{matrix} x^{2} + xy - 12y^{2} = 0\ \ \\ 2x^{2} - 3xy + y^{2} = 90 \\ \end{matrix}\text{\ \ \ \ } \right.\ \]

\[x^{2} + y(x) - 12y^{2} = 0\]

\[D = y^{2} + 4 \cdot 12y^{2} = 49y^{2}\]

\[x_{1} = \frac{- y + 7y}{2} = 3y;\ \ \ \]

\[x_{2} = \frac{- y - 7y}{2} = - 4y.\]

\[Подставим\ во\ второе\ \]

\[уравнение.\]

\[1)\ x_{1} = 3y:\]

\[2 \cdot 9y^{2} - 9y^{2} + y^{2} = 90\]

\[18y^{2} - 8y^{2} = 90\]

\[10y^{2} = 90\]

\[y^{2} = 9\]

\[y = \pm 3.\]

\[x = 3 \cdot ( \pm 3) = \pm 9.\]

\[2)\ x_{2} = - 4y:\]

\[2 \cdot 16y^{2} + 12y^{2} + y^{2} = 90\]

\[32y^{2} + {13y}^{2} = 90\]

\[45y^{2} = 90\]

\[y^{2} = 2\]

\[y = \pm \sqrt{2}.\]

\[x = - 4 \cdot \left( \pm \sqrt{2} \right) = \pm 4\sqrt{2}.\]

\[Ответ:\ \ \left( - 4\sqrt{2};\sqrt{2} \right);\ \ \]

\[\left( 4\sqrt{2};\ - \sqrt{2} \right);\ \ ( - 9; - 3);(9;3).\]