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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 467
Задание 467
\[\boxed{\text{467\ (467).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ \left\{ \begin{matrix} x + y - xy = 1 \\ x + y + xy = 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x + y - xy + x + y + xy = 10 \\ x + y - xy - x - y - xy = - 8 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2x + 2y = 10\ \ \ \ \ |\ :2 \\ - 2xy = - 8\ \ \ \ \ \ \ \ \ \ \ |\ :2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} x + y = 5 \\ xy = 4\ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x = 5 - y\]
\[5y - y^{2} - 4 = 0\]
\[y_{1} + y_{2} = 5,\ \ y_{1} = 4\]
\[y_{1}y_{2} = 4,\ \ y_{2} = 1\]
\[\left\{ \begin{matrix} y = 4 \\ x = 1 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} y = 1 \\ x = 4 \\ \end{matrix} \right.\ \]
\[Ответ:(1;4),\ (4;1).\]
\[2)\ \left\{ \begin{matrix} 3xy + 2x = - 4 \\ 3xy + y = - 8\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} 3xy + 2x - 3xy - y = 4 \\ 3xy + 2x = - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2x - y = 4\ \ \ \ \ \ \ \ \\ 3\text{xy} + 2x = - 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} y = 2x - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x \cdot (2x - 4) + 2x + 4 = 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = 2x - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 6x^{2} - 12x + 2x + 4 = 0\ \ \ \ |\ :2 \\ \end{matrix} \right.\ \]
\[3x^{2} - 5x + 2 = 0\]
\[D = 25 - 24 = 1\]
\[x_{1,2} = \frac{5 \pm 1}{6}\]
\[\left\{ \begin{matrix} x = 1\ \ \ \\ y = - 2 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \ \left\{ \begin{matrix} x = \frac{2}{3}\text{\ \ \ \ \ } \\ y = - \frac{8}{3} \\ \end{matrix} \right.\ \]
\[Ответ:(1;\ - 2);\ \left( \frac{2}{3};\ - \frac{8}{3} \right).\]
\[3)\ \left\{ \begin{matrix} xy - x = 24 \\ xy - y = 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} xy - x - xy + y = - 1 \\ xy - y = 25\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y - x = - 1 \\ xy - y = 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = x - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x(x - 1) - x + 1 - 25 = 0 \\ \end{matrix} \right.\ \]
\[x^{2} - x - x + 1 - 25 = 0\]
\[x^{2} - 2x - 24 = 0\]
\[x_{1} + x_{2} = 2,\ \ x_{1} = 6\]
\[x_{1}x_{2} = - 24,\ \ x_{2} = - 4\]
\[\left\{ \begin{matrix} x = 6 \\ y = 5 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x = - 4 \\ y = - 5 \\ \end{matrix} \right.\ \]
\[Ответ:(6;5),\ ( - 4;\ - 5).\]
\[4)\ \left\{ \begin{matrix} 2x^{2} + y^{2} = 66 \\ 2x^{2} - y^{2} = 34 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} 2x^{2} + y^{2} - 2x^{2} + y^{2} = 32 \\ 2x^{2} + y^{2} + 2x^{2} - y^{2} = 100 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2y^{2} = 32 \\ 4x^{2} = 100 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y^{2} = 16 \\ x^{2} = 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} y = \pm 4 \\ x = \pm 5 \\ \end{matrix} \right.\ \]
\[Ответ:(5;4),\ (5;\ - 4),\ ( - 5;4),\ \]
\[( - 5;\ - 4)\text{.\ }\]
\[\boxed{\text{467.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ y = \frac{8x + 2x^{2} - x^{3}}{x};\ \ \ x \neq 0\]
\[y = - x^{2} + 2x + 8;\ \ \]
\[a = - 1 < 0 - ветви\ вниз.\]
\[x_{0} = \frac{- 2}{- 2} = 1;\]
\[y_{0} = - 1 + 2 + 8 = 9;\]
\[вершина\ параболы - \text{\ \ }(1;9).\]
\[y = 0:\ \ \]
\[- x^{2} + 2x + 8 = 0\]
\[x_{1} + x_{2} = 2,\ \ x_{1} = 4\]
\[x_{1} \cdot x_{2} = - 8,\ \ x_{2} = - 2\]
\[( - 2;0),\ \ (4;0).\]
\[x = 0:\ \ \]
\[y = 8,\ \ (0;8).\]
\[y( - 1) = - 1 - 2 + 8 = 5.\]
\[2)\ y = \frac{x^{3} - 8}{x - 2} - 3\]
\[y = \frac{(x - 2)\left( x^{2} + 2x + 4 \right)}{(x - 2)} - 3 =\]
\[= x^{2} + 2x + 4 - 3;\ \ \ \ x \neq 2\]
\[y = x^{2} + 2x + 1\]
\[y = (x + 1)^{2}\]
\[a = 1 > 0 - ветви\ вверх;\]
\[( - 1;0) - вершина\ параболы.\]
\[3)\ y = \frac{x^{4} - 16}{x^{2} - 4}\]
\[y = \frac{\left( x^{2} - 4 \right)\left( x^{2} + 4 \right)}{x^{2} - 4};\ \ x \neq \pm 2\]
\[y = x^{2} + 4\]
\[a = 1 > 0 - ветви\ вверх.\]
\[(4;0) - вершина\ параболы.\]
\[4)\ y = \frac{x^{4} + 4x^{2} - 5}{x^{2} - 1}\]
\[y = \frac{\left( x^{2} - 1 \right)\left( x^{2} + 5 \right)}{\left( x^{2} - 1 \right)};\ \ x \neq \pm 1\]
\[y = x^{2} + 5\]
\[a = 1 > 0 - ветви\ вверх;\]
\[(0;5) - вершина\ параболы.\]
