Решебник по алгебре 9 класс Макарычев Задание 1070

 

\[\boxed{\text{1070\ (1070).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[= a_{1} + 3d,\]

\[d - разность\ арифметической\ \]

\[прогрессии.\]

\[По\ условию:\]

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

\[a_{2} + 2d =\]

\[= a_{2}^{2} + \left( a_{2} - d \right)^{2} + \left( a_{2} + d \right)^{2}\]

\[3a_{2}^{2} + 2d^{2} - a_{2} - 2d = 0\]

\[2d^{2} - 2d + \left( 3a_{2}^{2} - a_{2} \right) = 0\]

\[D = 4 - 4 \cdot 2 \cdot \left( 3a_{2}^{2} - a_{2} \right) =\]

\[= 4 - 24a_{2}^{2} + 8a_{2} \Longrightarrow D \geq 0,\]

\[6a_{2}^{2} - 2a_{2} - 1 \leq 0\]

\[D = 4 + 24 = 28,\]

\[a_{2} = \frac{2 \pm \sqrt{28}}{12} = \frac{1 \pm \sqrt{7}}{6},\]

\[\frac{1 - \sqrt{7}}{6} \leq a_{2} \leq \frac{1 + \sqrt{7}}{6} \Longrightarrow a_{2} =\]

\[= 0 - так\ как\ целое\ число.\]

\[2d^{2} - 2d = 0\]

\[d(d - 1) = 0\]

\[d = 0 \Longrightarrow не\ подходит\ по\ \]

\[условию.\]

\[d = 1.\]

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

\[a_{2} = 0\]

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

\[a_{4} = a_{2} + 2d = 2\]

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

\[= ( - 1)^{2} + 0^{2} + 1^{2} \Longrightarrow верно.\]

\[Ответ:a_{1} = - 1,\ a_{2} = 0,\ a_{3} = 1,\]

\[a_{4} = 2.\]