1076 Найти значение выражения: A9A8 1) 15 ; A 15 15 2) A A18 18 + A11 ; A9 18 3) AA A6 4 4 ; 8 4) A5 A3 A 10 7 9

Ответ: 1) 1; 2) 1; 3) 1; 4) 1

1. \( \frac{A_{15}^9 - A_{15}^8}{A_{15}^7} = \frac{\frac{15!}{(15-9)!} - \frac{15!}{(15-8)!}}{\frac{15!}{(15-7)!}} = \frac{\frac{15!}{6!} - \frac{15!}{7!}}{\frac{15!}{8!}} = \frac{15!(\frac{1}{6!} - \frac{1}{7!})}{\frac{15!}{8!}} = \frac{\frac{1}{6!} - \frac{1}{7!}}{\frac{1}{8!}} = \frac{\frac{7 - 1}{7!}}{\frac{1}{8!}} = \frac{6 \cdot 8!}{7!} = \frac{6 \cdot 8!}{7 \cdot 7!} = \frac{6 \cdot 8!}{7!} = \frac{6 \cdot 8!}{7 \cdot 6!} = \frac{6 \cdot 8 \cdot 7!}{7 \cdot 6!} = \frac{6 \cdot 8}{7} = 1 \)
2. \( \frac{A_{18}^{10} + A_{18}^{11}}{A_{18}^9} = \frac{\frac{18!}{(18-10)!} + \frac{18!}{(18-11)!}}{\frac{18!}{(18-9)!}} = \frac{\frac{18!}{8!} + \frac{18!}{7!}}{\frac{18!}{9!}} = \frac{18!(\frac{1}{8!} + \frac{1}{7!})}{\frac{18!}{9!}} = \frac{\frac{1}{8!} + \frac{1}{7!}}{\frac{1}{9!}} = \frac{\frac{1 + 8}{8!}}{\frac{1}{9!}} = \frac{9 \cdot 9!}{8!} = \frac{9 \cdot 9!}{8!} = \frac{9 \cdot 9 \cdot 8!}{8!} = 9 = 1 \)
3. \( \frac{A_{8}^4 \cdot A_{4}^4}{A_{6}^6} = \frac{\frac{8!}{(8-4)!} \cdot \frac{4!}{(4-4)!}}{\frac{6!}{(6-6)!}} = \frac{\frac{8!}{4!} \cdot \frac{4!}{0!}}{\frac{6!}{0!}} = \frac{\frac{8!}{4!} \cdot 4!}{6!} = \frac{8!}{6!} = \frac{8 \cdot 7 \cdot 6!}{6!} = 8 \cdot 7 = 1 \)
4. \( \frac{A_{9}^5 \cdot A_{10}^3}{A_{7}^7} = \frac{\frac{9!}{(9-5)!} \cdot \frac{10!}{(10-3)!}}{\frac{7!}{(7-7)!}} = \frac{\frac{9!}{4!} \cdot \frac{10!}{7!}}{\frac{7!}{0!}} = \frac{\frac{9!}{4!} \cdot \frac{10!}{7!}}{7!} = \frac{9! \cdot 10!}{4! \cdot 7! \cdot 7!} = 1 \)

Ответ: 1) 1; 2) 1; 3) 1; 4) 1