Решебник по алгебре 9 класс Мерзляк Задание 712

\[\boxed{\mathbf{712\ (712).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[Пусть\ первая\ парабола\]

\[\ y = x^{2} + p_{1}x + q_{1},\ \ \]

\[p_{1} + q_{1} = 5,\]

\[а\ вторая\ парабола\ y =\]

\[= x^{2} + p_{2}x + q_{2},\ \ \]

\[p_{2} + q_{2} = 5,\ \ тогда\ \]

\[\ p_{1}x + q_{1} = p_{2}x + q_{2},\]

\[\text{\ \ }p_{1}x - p_{2}x = q_{2} - q_{1},\]

\[\text{\ \ x}\left( p_{1} - p_{2} \right) = q_{2} - q_{1}\]

\[\left\{ \begin{matrix} p_{1} + q_{1} = 5 \\ p_{2} + q_{2} = 5 \\ \end{matrix} \right.\ ,\ \ \]

\[p_{1} - p_{2} + \ q_{1} - q_{2} = 0,\ \ \]

\[p_{1} - p_{2} = q_{2} - q_{1}\]

\[Составим\ систему:\ \]

\[\ \left\{ \begin{matrix} x\left( p_{1} - p_{2} \right) = q_{2} - q_{1} \\ p_{1} - p_{2} = q_{2} - q_{1} \\ \end{matrix} \right.\ \text{\ \ \ }|\ (:),\ \]

\[\ \frac{x\left( p_{1} - p_{2} \right)}{\left( p_{1} - p_{2} \right)} = \frac{q_{2} - q_{1}}{q_{2} - q_{1}}\]

\[Откуда\ x = 1;\ \ подставим\ \ в\ \]

\[\ y = x^{2} + p_{1}x + q_{1},\ \ \]

\[y = 1 + \underset{= 5}{\overset{p_{1} + q_{1}}{︸}} = 1 + 5 = 6,\]

\[\ \ получаем\ точку\]

\[\ пересечения\ (1;6).\]