Решебник по алгебре 9 класс Мерзляк Задание 752

\[\boxed{\mathbf{752\ (752).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[a_{1} = y^{2} - 2y,\ \ a_{2} = 3y + 5,\]

\[\text{\ \ }a_{3} = 4y + 13,\ \ \]

\[a_{4} = 2y^{2} - y + 25\]

\[a_{1} + a_{4} = a_{3} + a_{2}\]

\[y^{2} - 2y + 2y^{2} - y + 25 =\]

\[= 3y + 5 + 4y + 13\]

\[3y^{2} - 10y + 7 = 0\]

\[D = 100 - 84 = 16\]

\[y_{1} = \frac{10 + 4}{6} = \frac{7}{3},\]

\[\text{\ \ }y_{2} = \frac{10 - 4}{6} = 1\]

\[При\ y = \frac{7}{3},\ \ \]

\[a_{1} = \frac{49}{9} - \frac{14}{3} = \frac{7}{9},\]

\[\text{\ \ }a_{2} = 7 + 5 = 12,\ \]

\[a_{3} = \frac{28}{3} + 13 = \frac{67}{3},\]

\[\text{\ \ }a_{4} = \frac{2 \cdot 49}{9} - \frac{7}{3} + 25 = \frac{302}{9}\]

\[Проверим:a_{2} - a_{1} \neq a_{3} -\]

\[- a_{2} \neq a_{4} - a_{3},\ \ \]

\[следовательно,\ y = \frac{7}{3}\ \Longrightarrow не\ \]

\[удовлетворяет.\]

\[При\ y = 1:\ \ \]

\[a_{1} = 1 - 2 = - 1,\ \ \]

\[a_{2} = 3 + 5 = 8,\ \ \]

\[a_{3} = 4 + 13 = 17,\]

\[a_{4} = 2 - 1 + 25 = 26.\]

\[Ответ:при\ y = 1 \Longrightarrow \ a_{1} =\]

\[= - 1,a_{2} = 8,\ a_{3} = 17,\ a_{4} = 26.\ \]