Решебник по алгебре 8 класс Макарычев ФГОС Задание 1329

\[\boxed{\text{1329.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} 2x^{2} + 4xy - 5y = 1 \\ x^{2} + xy - 6y^{2} = 0\ \ \\ \end{matrix} \right.\ \]

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

\[D = y^{2} + 24y^{2} = 25y^{2}\]

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

\[x_{2} = \frac{- y + 5y}{2} = 2y.\]

\[1)\ x = - 3y:\]

\[2 \cdot ( - 3y)^{2} + 4y \cdot ( - 3y) -\]

\[- 5y = 0\]

\[18y^{2} - 12y^{2} - 5y = 0\]

\[6y^{2} - 5y = 0\]

\[6y\left( y - \frac{5}{6} \right) = 0\]

\[y = 0;\ \ y = \frac{5}{6}.\]

\[x = 0;\ \ \ x = - \frac{5}{2} = - 2,5.\]

\[2)\ x = 2y:\]

\[2 \cdot (2y)^{2} + 4y \cdot 2y - 5y = 0\]

\[8y^{2} + 8y^{2} - 5y = 0\]

\[16y^{2} - 5y = 0\]

\[16y\left( y - \frac{5}{16} \right) = 0\]

\[y = 0;\ \ y = \frac{5}{16}.\]

\[x = 0;\ \ y = \frac{5}{8}.\]

\[Ответ:(0;0);\ \ \left( \frac{5}{6};\ - \frac{5}{2} \right);\]

\[\left( \frac{5}{16};\frac{5}{8} \right).\]

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

\[\left\{ \begin{matrix} 3x^{2} - 2xy - y^{2} = 0\ \ \ \ \ \ \ \\ 3y^{2} - 2x^{2} - 4x + 3 = 0 \\ \end{matrix} \right.\ \]

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

\[D_{1} = y^{2} + 3y^{2} = 4y^{2}\]

\[x_{1} = \frac{y + 2y}{3} = y.\]

\[x_{2} = \frac{y - 2y}{3} = - \frac{y}{3}.\]

\[1)\ x = y:\]

\[3y^{2} - 2y^{2} - 4y + 3 = 0\]

\[y^{2} - 4y + 3 = 0\]

\[D_{1} = 4 - 3 = 1\]

\[y_{1} = 2 + 1 = 3;\ \ \ \]

\[y_{2} = 2 - 1 = 1.\]

\[x_{1} = 3;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{2} = 1.\]

\[2)\ x = - \frac{y}{3}:\]

\[3y^{2} - 2 \cdot \left( - \frac{y}{3} \right)^{2} -\]

\[- 4 \cdot \left( - \frac{y}{3} \right) + 3 = 0\]

\[3y^{2} - \frac{2y^{2}}{9} + \frac{4y}{3} + 3 = 0\ \ \ \ | \cdot 9\]

\[27y^{2} - 2y^{2} + 12y + 27 = 0\]

\[25y^{2} + 12y + 27 = 0\]

\[D_{1} = 36 - 25 \cdot 27 < 0\]

\[нет\ корней.\]

\[Ответ:(3;3);(1;1).\]