ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 441

\[1)\ A_{m}^{2} = 90:\]

\[\frac{m!}{(m - 2)!} = 90\]

\[\frac{m(m - 1)(m - 2)!}{(m - 2)!} = 90\]

\[m(m - 1) = 90\]

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

\[D = 1 + 360 = 361\]

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

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

\[Ответ:\ \ 10.\]

\[2)\ A_{m}^{3} = 56m:\]

\[\frac{m!}{(m - 3)!} = 56m\]

\[\frac{m(m - 1)(m - 2)(m - 3)!}{(m - 3)!} = 56m\]

\[m(m - 1)(m - 2) = 56m\]

\[m\left( m^{2} - 3m + 2 \right) = 56m\]

\[m\left( m^{2} - 3m - 54 \right) = 0\]

\[D = 9 + 216 = 225\]

\[m_{1} = \frac{3 - 15}{2} = - 6;\]

\[m_{2} = \frac{3 + 15}{2} = 9.\]

\[Ответ:\ \ 9.\]

\[3)\ A_{m + 1}^{2} = 156:\]

\[\frac{(m + 1)!}{(m + 1 - 2)!} = 156\]

\[\frac{(m + 1)m(m - 1)!}{(m - 1)!} = 156\]

\[m(m + 1) = 156\]

\[m^{2} + m - 156 = 0\]

\[D = 1 + 624 = 625:\]

\[m_{1} = \frac{- 1 - 25}{2} = - 13;\]

\[m_{2} = \frac{- 1 + 25}{2} = 12.\]

\[Ответ:\ \ 12.\]

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

\[\frac{m!}{(m - 5)!} = 18 \bullet \frac{(m - 2)!}{(m - 2 - 4)!}\]

\[\frac{m(m - 1)(m - 2)!}{(m - 5)(m - 6)!} = 18 \bullet \frac{(m - 2)!}{(m - 6)!}\]

\[\frac{m(m - 1)}{m - 5} = 18\]

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

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

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

\[D = 361 - 360 = 1\]

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

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

\[Ответ:\ \ 9;\ 10.\]