Решебник по алгебре 8 класс Макарычев ФГОС Задание 1326

\[\boxed{\text{1326.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} - 3y^{2} - y = - 6 \\ 2x^{2} - 3y^{2} = - 4\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} x^{2} = \left( 3y^{2} + y - 6 \right)\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 2 \cdot \left( 3y^{2} + y - 6 \right) - 3y^{2} = - 4 \\ \end{matrix} \right.\ \]

\[6y^{2} + 2y - 12 - 3y^{2} + 4 = 0\]

\[3y^{2} + 2y - 8 = 0\]

\[D_{1} = 1 + 24 = 25\]

\[y_{1} = \frac{- 1 + 5}{3} = \frac{4}{3};\ \ \ \]

\[y_{2} = \frac{- 1 - 5}{3} = - 2.\]

\[1)\ y = \frac{4}{3}:\]

\[x^{2} = 3 \cdot \frac{16}{9} + \frac{4}{3} - 6 = \frac{16}{3} +\]

\[+ \frac{4}{3} - 6 = \frac{20}{3} - \frac{18}{3} = \frac{2}{3}\]

\[x = \pm \sqrt{\frac{2}{3}}.\]

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

\[x^{2} = 3 \cdot 4 - 2 - 6 = 4\]

\[x = \pm 2.\]

\[Ответ:\left( - \sqrt{\frac{2}{3}};\frac{4}{3} \right);\ \ \left( \sqrt{\frac{2}{3}};\frac{4}{3} \right);\]

\[( - 2; - 2);(2; - 2).\]

\[\textbf{б)}\ \left\{ \begin{matrix} 2x^{2} + xy = 16\ \ \ \ \ \ \ \ \ \\ 3x^{2} + xy - x = 18\ \\ \end{matrix} \right.\ ( - )\]

\[- x^{2} + x = - 2\]

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

\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 2\]

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

\[xy = 16 - 2x^{2}\]

\[y = \frac{16 - 2x²}{x}\ \]

\[y(2) = \frac{16 - 8}{2} = 4;\]

\[y( - 1) = \frac{16 - 2}{- 1} = - 14.\]

\[Ответ:(2;4);( - 1; - 14).\]