Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1077

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Год:2020-2021-2022-2023
Тип:учебник
Серия:Базовый и углубленный уровни

Задание 1077

\[\boxed{\mathbf{1077}\mathbf{.}}\]

\[1)\ A_{m}^{2} = 72:\]

\[m \bullet (m - 1) = 72\]

\[m^{2} - m - 72 = 0\]

\[D = 1^{2} + 4 \bullet 72 = 1 + 288 =\]

\[= 289\]

\[m_{1} = \frac{1 - 17}{2} = - 8\ \ и\ \]

\[m_{2} = \frac{1 + 17}{2} = 9.\]

\[Ответ:\ \ m = 9.\]

\[2)\ A_{m}^{2} = 56:\]

\[m \bullet (m - 1) = 56\]

\[m^{2} - m - 56 = 0\]

\[D = 1^{2} + 4 \bullet 56 = 1 + 224 = 225\]

\[m_{1} = \frac{1 - 15}{2} = - 7\ \ и\ \ \]

\[m_{2} = \frac{1 + 15}{2} = 8.\]

\[Ответ:\ \ m = 8.\]

\[3)\ A_{m}^{3} = 12m:\]

\[m \bullet (m - 1) \bullet (m - 2) = 12m\]

\[m^{2} - 2m - m + 2 - 12 = 0\]

\[m^{2} - 3m - 10 = 0\]

\[D = 3^{2} + 4 \bullet 10 = 9 + 40 = 49\]

\[m_{1} = \frac{3 - 7}{2} = - 2\ \ и\ \]

\[\ m_{2} = \frac{3 + 7}{2} = 5.\]

\[Ответ:\ \ m = 5.\]

\[4)\ A_{m}^{3} = 20m:\]

\[m \bullet (m - 1) \bullet (m - 2) = 20m\]

\[m^{2} - 2m - m + 2 - 20 = 0\]

\[m^{2} - 3m - 18 = 0\]

\[D = 3^{2} + 4 \bullet 18 = 9 + 72 = 81\]

\[m_{1} = \frac{3 - 9}{2} = - 3\ \ и\ \ \]

\[m_{2} = \frac{3 + 9}{2} = 6.\]

\[Ответ:\ \ m = 6.\]

\[5)\ A_{m + 1}^{2} = 110:\]

\[(m + 1) \bullet m = 110\]

\[m^{2} + m - 110 = 0\]

\[D = 1^{2} + 4 \bullet 110 = 1 + 440 =\]

\[= 441\]

\[m_{1} = \frac{- 1 - 21}{2} = - 11\ \ и\ \ \]

\[m_{2} = \frac{- 1 + 21}{2} = 10.\]

\[Ответ:\ \ m = 10.\]

\[6)\ A_{m + 2}^{2} = 90:\]

\[(m + 2) \bullet (m + 1) = 90\]

\[m^{2} + m + 2m + 2 - 90 = 0\]

\[m^{2} + 3m - 88 = 0\]

\[D = 3^{2} + 4 \bullet 88 = 9 + 352 = 361\]

\[m_{1} = \frac{- 3 - 19}{2} = - 11\ \ и\ \ \]

\[m_{2} = \frac{- 3 + 19}{2} = 8.\]

\[Ответ:\ \ m = 8.\]

\[7)\ A_{m}^{5} = 18A_{m - 2}^{4}:\]

\[m(m - 1) = 18(m - 5)\]

\[m^{2} - m = 18m - 90\]

\[m^{2} - 19m + 90 = 0\]

\[D = 19^{2} - 4 \bullet 90 =\]

\[= 361 - 360 = 1\]

\[m_{1} = \frac{19 - 1}{2} = 9\ \ и\ \ \]

\[m_{2} = \frac{19 + 1}{2} = 10.\]

\[Ответ:\ \ m_{1} = 9;\ \ m_{2} = 10.\]

\[8)\ (m - 4) \bullet A_{m}^{4} =\]

\[= 21(m - 5) \bullet A_{m - 2}^{3}:\]

\[m(m - 1) = 21(m - 5)\]

\[m^{2} - m = 21m - 105\]

\[m^{2} - 22m + 105 = 0\]

\[D = 22^{2} - 4 \bullet 105 =\]

\[= 484 - 420 = 64\]

\[m_{1} = \frac{22 - 8}{2} = 7\ \ и\ \ \]

\[m_{2} = \frac{22 + 8}{2} = 15.\]

\[Ответ:\ \ m_{1} = 7;\ \ m_{2} = 15.\]

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