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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 466
Задание 466
\[\boxed{\text{466\ (466).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ \left\{ \begin{matrix} 3y - 2xy = 2 \\ x + 2xy = 5\ \ \\ \end{matrix} \right.\ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]
\[\ \left\{ \begin{matrix} 3y - 2xy + x + 2xy = 7 \\ x + 2xy = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 3y + x = 7 \\ x + 2xy = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 7 - 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 7 - 3y + 2y \cdot (7 - 3y) = 5 \\ \end{matrix} \right.\ \]
\[7 - 3y + 14y - 6y^{2} - 5 = 0\]
\[6y^{2} - 11y - 2 = 0\]
\[D = 121 + 48 = 169\]
\[y_{1,2} = \frac{11 \pm 13}{12}\]
\[y_{1} = 2\]
\[y_{2} = - \frac{1}{6}\]
\[\left\{ \begin{matrix} y = 2 \\ x = 1 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \ \left\{ \begin{matrix} y = - \frac{1}{6} \\ x = 7,5 \\ \end{matrix} \right.\ \]
\[Ответ:(1;2);\left( 7,5;\ - \frac{1}{6} \right).\]
\[2)\ \left\{ \begin{matrix} xy + y = 30 \\ xy + x = 28 \\ \end{matrix} - \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} xy + y - xy - x = 2 \\ xy + x = 28\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y - x = 2 \\ xy + x = 28 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = 2 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \cdot (2 + x) + x = 28 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} y = 2 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x + x^{2} + x - 28 = 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = 2 + x \\ x^{2} + 3x - 28 = 0 \\ \end{matrix} \right.\ \]
\[x_{1} + x_{2} = - 3,\ \ x_{1} = - 7\]
\[x_{1}x_{2} = - 28,\ \ x_{2} = 4\]
\[\left\{ \begin{matrix} x = - 7 \\ y = - 5 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x = 4 \\ y = 6 \\ \end{matrix} \right.\ \]
\[Ответ:( - 7;\ - 5);\ (4;6).\]
\[x = 2 + y\]
\[2 \cdot (2 + y)^{2} - 5y \cdot (2 + y) -\]
\[- 2 \cdot (2 + y) + 3y = 0\]
\[8 + 8y + 2y^{2} - 10y - 5y^{2} -\]
\[- 4 - 2y + 3y = 0\]
\[- 3y^{2} - y + 4 = 0\]
\[D = 1 + 48 = 49\]
\[y_{1,2} = \frac{1 \pm 7}{- 6}\]
\[\left\{ \begin{matrix} y = - \frac{4}{3} \\ x = \frac{2}{3} \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} y = 1 \\ x = 3 \\ \end{matrix} \right.\ \]
\[Ответ:\left( \frac{2}{3};\ - \frac{4}{3} \right);\ (3;1)\text{.\ }\]
