Решебник по алгебре 9 класс Мерзляк Задание 1031

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Год:2023
Тип:учебник
Серия:Алгоритм успеха

Задание 1031

\[\boxed{\mathbf{1031\ (1031).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\left\{ \begin{matrix} a_{3} + a_{5} + a_{8} = 18 \\ a_{2} + a_{4} = - 2\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} a_{1} + 2d + a_{1} + 4d + a_{1} + 7d = 18 \\ a_{1} + d + a_{1} + 3d = - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 3a_{1} + 13d = 18\ \ \ \ | \cdot 2 \\ 2a_{1} + 4d = - 2\ \ \ \ \ | \cdot 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 6a_{1} + 26d = 36 \\ 6a_{1} + 12d = - 6 \\ \end{matrix} \right.\ \ - \ \ \ \]

\[\ \left\{ \begin{matrix} 14d = 42\ \ \ \ \ \ \ \ \\ a_{1} = \frac{- 2 - 4d}{2} \\ \end{matrix}\ \right.\ \]

\[\left\{ \begin{matrix} d = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a_{1} = \frac{- 2 - 12}{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} d = 3\ \ \ \ \ \\ a_{1} = - 7 \\ \end{matrix} \right.\ \]

\[Ответ:d = 3;\ \ a_{1} = - 7.\]

\[2)\ \left\{ \begin{matrix} a_{5} - a_{3} = - 4 \\ a_{2} \cdot a_{4} = - 3\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a_{1} + 4d - a_{1} - 2d = - 4 \\ \left( a_{1} + d \right)\left( a_{1} + 3d \right) = - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} 2d = - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left( a_{1} + d \right)\left( a_{1} + 3d \right) = - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} d = - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left( a_{1}–2 \right)\left( a_{1} - 6 \right) = - 3 \\ \end{matrix} \right.\ \]

\[a^{2} - 8a + 12 + 3 = 0\]

\[a^{2} - 8a + 15 = 0\]

\[a_{1} + a_{2} = 8,\ \ a_{1} = 5\]

\[a_{1}a_{2} = 15,\ \ a_{2} = 3\]

\[Ответ:d = - 2;\ a_{1} = - 5\ \ или\]

\[\text{\ \ }a_{1} = 3.\]

\[3)\ \left\{ \begin{matrix} a_{2} + a_{4} + a_{6} = 36 \\ a_{2} \cdot a_{3} = 54\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} a_{1} + d + a_{1} + 3d + a_{1} + 5d = 36 \\ \left( a_{1} + d \right)\left( a_{1} + 2d \right) = 54\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 3a_{1} + 9d = 36\ \ \ \ \ \ \ \ \ \ |\ \ :3 \\ \left( a_{1} + d \right)\left( a_{1} + 2d \right) = 54 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} a_{1} + 3d = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left( a_{1} + d \right)\left( a_{1} + 2d \right) = 54 \\ \end{matrix} \right.\ ,\ \ \]

\[a_{1} = 12 - 3d\]

\[(12 - 3d + d)(12 - 3d + 2d) =\]

\[= 54\]

\[(12 - 2d)(12 - d) = 54\ \ \ \ |\ :2\]

\[(6 - d)(12 - d) = 27\]

\[72 - 18d + d^{2} - 27 = 0\]

\[d^{2} - 18d + 45 = 0\]

\[d_{1} + d_{2} = 18,\ \ d = 3\]

\[d_{1}d_{2} = 45,\ \ d = 15\]

\[d = 3,\ \ a_{1} = 12 - 3 \cdot 3 = 3\]

\[d = 15,\ \ \]

\[a_{1} = 12 - 3 \cdot 15 = - 33\]

\[Ответ:d = 3;\ a_{1} = 3\ \ или\ \]

\[\ d = 15;\ a_{1} = - 33.\]

\[4)\ \left\{ \begin{matrix} S_{5} - S_{2} - a_{5} = 0,1 \\ a_{7} + S_{4} = 0,1\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[S_{5} - S_{2} - a_{5} = \frac{\left( a_{1} + a_{5} \right)}{2} \cdot 5 -\]

\[- \left( a_{1} + a_{1} \right) - a_{5} = 5a_{1} +\]

\[+ 5a_{5} - 2a_{1} - 2a_{2} - 2a_{5} =\]

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

\[+ 3 \cdot \left( a_{1} + 4d \right) - 2 \cdot \left( a_{1} + d \right) =\]

\[= 3a_{1} + 3a_{1} + 12d - 2a_{1} -\]

\[- 2d = 4a_{1} + 10d\]

\[4a_{1} + 10d = 0,1 \cdot 2\ \ \ \ \ |\ :2\]

\[2a_{1} + 5d = 0,1\]

\[a_{7} + S_{4} = a_{7} + \frac{a_{1} + a_{4}}{2} \cdot 4 =\]

\[= a_{1} + 6d + 2a_{1} + 2a_{4} =\]

\[= a_{1} + 6d + 2a_{1} +\]

\[+ 2 \cdot \left( a_{1} + 3d \right) = a_{1} + 6d +\]

\[+ 2a_{1} + 2a_{1} + 6d =\]

\[= 5a_{1} + 12d = 0,1\]

\[\left\{ \begin{matrix} 2a_{1} + 5d = 0,1\ \ \\ 5a_{1} + 12d = 0,1 \\ \end{matrix} \right.\ - ,\ \ \]

\[- 3a_{1} - 7d = 0,\ \ \]

\[- 3a_{1} = 7d,\ \ a_{1} = - \frac{7}{3}d\]

\[2 \cdot \left( - \frac{7}{3}d \right) + 5d = 0,1\ \]

\[- \frac{14}{3}d + 5d = 0,1\]

\[- 14d + 15d = 0,3\]

\[d = 0,3,\ \ \]

\[a_{1} = - \frac{7}{3} \cdot 0,3 = - 0,7\]

\[Ответ:d = 0,3;\ \ a_{1} = - 0,7.\]

\[5)\ S_{4} = 9\ \ \ и\ \ \ S_{6} = 22,5\]

\[S_{4} = \frac{2a_{1} + 3d}{2} \cdot 4 =\]

\[= 2 \cdot \left( 2a_{1} + 3d \right) = 9\]

\[S_{6} = \frac{2a_{1} + 5d}{2} \cdot 6 =\]

\[= 3 \cdot \left( 2a_{1} + 5d \right) = 22,5\]

\[\left\{ \begin{matrix} 2 \cdot \left( 2a_{1} + 3d \right) = 9\ \ \ \ \ \ \ \ \ |\ :2 \\ 3 \cdot \left( 2a_{1} + 5d \right) = 22,5\ \ \ |\ :3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} 2a_{1} + 3d = 4,5 \\ 2a_{1} + 5d = 7,5 \\ \end{matrix} \right.\ -\]

\[- 2d = - 3,\ \ d = 1,5\]

\[2a_{1} + 3 \cdot 1,5 = 4,5\]

\[2a_{1} = 0,\ \ a_{1} = 0\]

\[Ответ:d = 1,5;\ \ \ a_{1} = 0.\]

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