ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 454

\[1)\ C_{m}^{3} = \frac{4}{15}C_{m + 2}^{4}\]

\[\frac{m!}{3!(m - 3)!} = \frac{4}{15} \bullet \frac{(m + 2)!}{4!(m - 2)!}\]

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

\[= \frac{4}{15} \bullet \frac{(m + 2)(m + 1)m!}{4 \bullet 3!(m - 2)(m - 3)!}\]

\[\frac{4}{15} \bullet \frac{(m + 2)(m + 1)}{4(m - 2)} = 1\]

\[(m + 2)(m + 1) = 15(m - 2)\]

\[m^{2} + m + 2m + 2 = 15m - 30\]

\[m^{2} - 12m + 32 = 0\]

\[D = 144 - 128 = 16\]

\[m_{1} = \frac{12 - 4}{2} = 4;\text{\ \ }\]

\[m_{2} = \frac{12 + 4}{2} = 8.\]

\[Ответ:\ \ 4;\ 8.\]

\[2)\ 12C_{m + 3}^{m - 1} = 55A_{m + 1}^{2}\]

\[12 \bullet \frac{(m + 3)!}{(m - 1)! \bullet 4!} = 55 \bullet \frac{(m + 1)!}{(m - 1)!}\]

\[12 \bullet \frac{(m + 3)(m + 2)(m + 1)!}{4 \bullet 3 \bullet 2} =\]

\[= 55 \bullet (m + 1)!\]

\[\frac{(m + 3)(m + 2)}{2} = 55\]

\[(m + 3)(m + 2) = 110\]

\[m^{2} + 2m + 3m + 6 = 110\]

\[m^{2} + 5m - 104 = 0\]

\[D = 25 + 416 = 441\]

\[m_{1} = \frac{- 5 - 21}{2} = - 13;\]

\[m_{2} = \frac{- 5 + 21}{2} = 8.\]

\[Ответ:\ \ 8.\]

\[3)\ 5C_{m}^{3} = C_{m + 2}^{4}\]

\[5 \bullet \frac{m!}{3!(m - 3)!} = \frac{(m + 2)!}{4!(m - 2)!}\]

\[5 \bullet \frac{m!}{3!(m - 3)!} =\]

\[= \frac{(m + 2)(m + 1)m!}{4 \bullet 3!(m - 2)(m - 3)!}\]

\[5 = \frac{(m + 2)(m + 1)}{4(m - 2)}\]

\[5 \bullet 4(m - 2) = (m + 2)(m + 1)\]

\[20m - 40 = m^{2} + 2m + m + 2\]

\[m^{2} - 17m + 42 = 0\]

\[D = 289 - 168 = 121\]

\[m_{1} = \frac{17 - 11}{2} = 3;\text{\ \ }\]

\[m_{2} = \frac{17 + 11}{2} = 14.\]

\[Ответ:\ \ 3;\ 14.\]

\[4)\ C_{3m + 1}^{3m - 1} = 120\]

\[\frac{(3m + 1)!}{(3m - 1)! \bullet 2!} = 120\]

\[\frac{(3m + 1)3m(3m - 1)!}{(3m - 1)! \bullet 2} = 120\]

\[(3m + 1) \bullet 3m = 120 \bullet 2\]

\[9m^{2} + 3m - 240 = 0\]

\[3m^{2} + m - 80 = 0\]

\[D = 1 + 960 = 961\]

\[m_{1} = \frac{- 1 - 31}{2 \bullet 3} = - \frac{16}{3};\text{\ \ }\]

\[m_{2} = \frac{- 1 + 31}{2 \bullet 3} = 5;\]

\[Ответ:\ \ 5.\]