Решебник по алгебре 9 класс Макарычев Задание 1061

\[\boxed{\text{1061\ (1061).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x³ - 9x^{2} + mx - 15 = 0\]

\[Используя\ теорему\ Виета,\ \]

\[составим\ систему\ уравнений:\]

\(\left\{ \begin{matrix} x_{1} + x_{2} + x_{3} = 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1} = m \\ x_{1} \cdot x_{2} \cdot x_{3} = 15\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \).

\[По\ условию,\ числа\ x_{1},x_{2},x_{3}\ \]

\[входят\ в\ арифметическую\ \]

\[прогрессию:\]

\[x_{2} = x_{1} + d,\ \ x_{3} = x_{1} + 2d.\]

\[x_{1} + x_{2} + x_{3} =\]

\[= x_{1} + x_{1} + d + x_{1} + 2d =\]

\[= 3x_{1} + 3d = 3x_{2} = 9 \Longrightarrow x_{2} = 3.\]

\[x_{1} \cdot x_{2} \cdot x_{3} =\]

\[= \left( x_{2} - d \right) \cdot x_{2} \cdot \left( x_{2} + d \right) =\]

\[= x_{2} \cdot \left( x_{2}^{2} - d^{2} \right) = 15,\]

\[x_{2}^{2} - d^{2} = 5\]

\[d^{2} = x_{2}^{2} - 5 = 9 - 5 = 4 \Longrightarrow\]

\[\Longrightarrow d = \pm 2.\]

\[\Longrightarrow x_{1} = 1,\ \ x_{2} = 3,\ \ \]

\[x_{3} = 5.\]

\[m = x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1} =\]

\[= 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 1 = 23.\]

\[Ответ:при\ m = 23.\]