Решебник по алгебре 9 класс Мерзляк Задание 465

\[\boxed{\text{465\ (465).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left\{ \begin{matrix} x^{3} - y^{3} = 56 \\ x - y = 2\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} (x - y)\left( x^{2} + xy + y^{2} \right) = 56 \\ x - y = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2x^{2} + 2xy + 2y^{2} = 56 \\ x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

\[+ y^{2} - 28 = 0\]

\[4 + 4y + y^{2} + 2y + y^{2} + y^{2} -\]

\[- 28 = 0\]

\[3y^{2} + 6y - 24 = 0\ \ \ \ \ \ \ \ \ \ \ \ \ |\ :3\]

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

\[y_{1} + y_{2} = - 2,\ \ y_{1} = - 4\]

\[y_{1}y_{2} = - 8,\ \ y_{2} = 2\]

\[\left\{ \begin{matrix} y = - 4 \\ x = - 2 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} y = 2 \\ x = 4 \\ \end{matrix} \right.\ \]

\[Ответ:( - 2;\ - 4),\ (4;2).\]

\[2)\ \left\{ \begin{matrix} 5x^{2} - y^{2} = - 4 \\ xy = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 5x^{2} - y^{2} = - 4 \\ x = \frac{3}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\frac{45}{y^{2}} - y^{2} + 4 = 0,\ \ y \neq 0\]

\[45 - y^{4} + 4y^{2} = 0\]

\[Пусть\ y^{2} = t,\ тогда:\]

\[- t^{2} + 4t + 45 = 0,\ \ t \geq 0\]

\[t_{1} + t_{2} = 4,\ \ t_{1} = 9\]

\[t_{1}t_{2} = - 45,\ \ t_{2} = - 5 -\]

\[не\ удовлетворяет.\]

\[y^{2} = 9\]

\[\left\{ \begin{matrix} y = 3 \\ x = 1 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \ \left\{ \begin{matrix} y = - 3 \\ x = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:(1;3),\ ( - 1;\ - 3)\text{.\ }\]