ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 737

\[1)\ \frac{A_{11}^{3} - A_{10}^{2}}{A_{9}^{1}} =\]

\[= \frac{\frac{11!}{(11 - 3)!} - \frac{10!}{(10 - 2)!}}{\frac{9!}{(9 - 1)!}} =\]

\[= \frac{\frac{11!}{8!} - \frac{10!}{8!}}{\frac{9!}{8!}} = \frac{11! - 10!}{9!} =\]

\[= \frac{11 \bullet 10! - 10!}{9!} = \frac{10 \bullet 10!}{9!} =\]

\[= \frac{10 \bullet 10 \bullet 9!}{9!} = 100.\]

\[2)\ \frac{A_{12}^{4} \bullet A_{7}^{7}}{A_{11}^{9}} = \frac{\frac{12!}{(12 - 4)!} \bullet \frac{7!}{(7 - 7)!}}{\frac{11!}{(11 - 9)!}} =\]

\[= \frac{\frac{12!}{8!} \bullet \frac{7!}{0!}}{\frac{11!}{2!}} = \frac{12 \bullet 11!}{8 \bullet 7!} \bullet \frac{7!}{1} \bullet \frac{2!}{11!} =\]

\[= \frac{12}{8} \bullet 2 = \frac{12}{4} = 3.\]

\[3)\ \frac{A_{6}^{3}}{P_{4}} + \frac{A_{11}^{6}}{11P_{6}} =\]

\[= \frac{6!}{(6 - 3)!}\ :4! + \frac{11!}{(11 - 6)!}\ :(11 \bullet 6!) =\]

\[= \frac{6 \bullet 5 \bullet 4!}{3! \bullet 4!} + \frac{11 \bullet 10 \bullet 9 \bullet 8 \bullet 7 \bullet 6!}{5! \bullet 11 \bullet 6!} =\]

\[= \frac{6 \bullet 5}{3 \bullet 2} + \frac{10 \bullet 9 \bullet 8 \bullet 7}{5 \bullet 4 \bullet 3 \bullet 2} =\]

\[= 5 + 3 \bullet 2 \bullet 7 = 5 + 42 = 47.\]

\[4)\ \left( \frac{C_{11}^{7}}{10} - \frac{C_{5}^{2}}{10} \right)\frac{P_{6}}{A_{6}^{4}} =\]

\[= \left( \frac{11 \bullet 9 \bullet 8}{4 \bullet 3 \bullet 2} - \frac{20}{20} \right) \bullet 2 =\]

\[= (11 \bullet 3 - 1) \bullet 2 =\]

\[= (33 - 1) \bullet 2 = 32 \bullet 2 = 64.\]