ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 1327

\[\boxed{\text{1327.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} 7xy + 2x^{2} - 4y^{2} = 0 \\ x^{2} - 5xy + y = - 11 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[2x^{2} + 8xy - 4y^{2} - xy = 0\]

\[2x(x + 4y) - y(x + 4y) = 0\]

\[(x + 4y)(2x - y) = 0\]

\[1)\ \left\{ \begin{matrix} x + 4y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5xy + y = - 11 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = - 4y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ ( - 4y)^{2} - 5( - 4y)y + y = - 11 \\ \end{matrix} \right.\ \]

\[16y^{2} + 20y^{2} + y + 11 = 0\]

\[36y^{2} + y + 11 = 0\]

\[D = 1 - 4 \cdot 36 \cdot 11 < 0\]

\[нет\ корней.\]

\[2)\ \left\{ \begin{matrix} 2x - y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5xy + y = - 11 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5 \cdot 2x \cdot x + 2x = - 11 \\ \end{matrix} \right.\ \]

\[x^{2} - 10x^{2} + 2x + 11 = 0\]

\[9x^{2} - 2x - 11 = 0\]

\[D_{1} = 1 + 99 = 100\]

\[x_{1} = \frac{1 + 10}{9} = \frac{11}{9};\ \ \ \ \ \]

\[x_{2} = \frac{1 - 10}{9} = - 1.\]

\[y_{1} = 2 \cdot \frac{11}{9} = \frac{22}{9};\ \ \ \ \ \]

\[\ y_{2} = 2 \cdot ( - 1) = - 2.\]

\[Ответ:\ \ ( - 1; - 2);\ \ \left( \frac{11}{9};\ \frac{22}{9} \right).\]

\[\textbf{б)}\ \left\{ \begin{matrix} 6x^{2} + 2xy - 3x - y = 0 \\ 2x^{2} - y^{2} + 2x + y = \frac{3}{2}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[6x^{2} + 2xy - 3x - y = 0\]

\[2x(3x + y) - (3x + y) = 0\]

\[(2x - 1)(3x + y) = 0.\]

\[1)\ 2x = 1\]

\[x = \frac{1}{2}\text{.\ \ \ \ \ }\]

\[Подставим:\]

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

\[2 \cdot \frac{1}{4} - y^{2} + 2 \cdot \frac{1}{2} + y = \frac{3}{2}\]

\[- y^{2} + y + \frac{1}{2} + 1 - \frac{3}{2} = 0\]

\[y^{2} - y = 0\]

\[y(y - 1) = 0\]

\[y = 0;\ \ y = 1.\]

\[2)\ 3x + y = 0\]

\[y = - 3x.\]

\[Подставим:\]

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

\[2x^{2} - ( - 3x)^{2} + 2x - 3x = \frac{3}{2}\]

\[2x^{2} - 9x^{2} - x - \frac{3}{2} = 0\]

\[- 7x^{2} - x - \frac{3}{2} = 0\ \ \ | \cdot ( - 2)\]

\[14x^{2} + 2x + 3 = 0\]

\[D_{1} = 1 - 42 = - 41 < 0\]

\[нет\ корней.\]

\[Ответ:(0,5;0);\ \ (0,5;1).\]