Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 315

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\[P(x) = x^{9} + bx^{8} + cx^{7}\]

\[P\left( - a_{1} \right) = \left( - a_{1} \right)^{9} +\]

\[+ 6 \cdot \left( - a_{1} \right)^{8} + c \cdot \left( - a_{1} \right)^{7}\]

\[p\left( - a_{2} \right) = \left( - a_{2} \right)^{9} +\]

\[+ 6 \cdot \left( - a_{2} \right)^{8} + c \cdot \left( - a_{2} \right)^{7}\]

\[\left\{ \begin{matrix} - a_{1}^{2} + ba_{1} - c = 0 \\ - a_{2}^{2} + ba_{2} - c = 0 \\ \end{matrix} \right.\ ;\ \]

\[\ \ c = - a_{1}^{2} + ba_{1}\]

\[- a_{2}^{2} + ba_{1} + a_{1}^{2} - ba_{1} = 0\]

\[\left( a_{2} - a_{1} \right)b = a_{2}^{2} - a_{1}^{2}\text{\ \ \ \ }\]

\[\ |\ :\left( a_{2} - a_{1} \right)\]

\[b = a_{2} + a_{1};\]

\[c = - a_{1}^{2} + a_{1}\left( a_{2} + a_{1} \right) =\]

\[= a_{1} \cdot a_{2}.\]

\[Что\ и\ требовалось\ доказать.\]