Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 427

\[k,\ m,\ n - простые\ числа.\]

\[1)\ \frac{19!}{18!} = \frac{19 \bullet 18!}{18!} = 19\]

\[2)\ \frac{22!}{20!} = \frac{22 \bullet 21 \bullet 20!}{20!} =\]

\[= 22 \bullet 21 = 462\]

\[3)\ \frac{6! \bullet 4!}{8!} = \frac{6! \bullet 4!}{8 \bullet 7 \bullet 6!} =\]

\[= \frac{4 \bullet 3 \bullet 2}{8 \bullet 7} = \frac{3}{7}\]

\[4)\ \frac{10!}{8! \bullet 3!} = \frac{10 \bullet 9 \bullet 8!}{8! \bullet 3!} = \frac{10 \bullet 9}{3 \bullet 2} =\]

\[= 5 \bullet 3 = 15\]

\[5)\ \frac{P_{n + 2}}{P_{n}} = \frac{(n + 2)!}{n!} =\]

\[= \frac{(n + 2)(n + 1)n!}{n!} =\]

\[= n^{2} + 3n + 2\]

\[6)\ \frac{P_{n + 1}}{P_{n + 3}} = \frac{(n + 1)!}{(n + 3)!} =\]

\[= \frac{(n + 1)!}{(n + 3)(n + 2)(n + 1)!} =\]

\[= \frac{1}{n^{2} + 5n + 6}\]

\[7)\ \frac{m! \bullet (m + 1)}{(m + 2)!} =\]

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

\[8)\ \frac{(k + 4)! \bullet (k + 5)}{(k + 6)!} =\]

\[= \frac{(k + 5)!}{(k + 6) \bullet (k + 5)!} = \frac{1}{k + 6}\]