Упр.737 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1. Найти значение выражения:
   
   1) $$A_{11}^{3}-\frac{A_{10}^{2}}{A_{9}^{1}}$$;
   
   2) $$A_{12}^{4}\cdot\frac{A_{7}^{7}}{A_{11}^{9}}$$;
   
   3) $$\frac{A_{6}^{3}}{P_{4}}+\frac{A_{11}^{6}}{11P_{6}}$$;
   
   4) $$\left(\frac{C_{11}^{7}}{10}-\frac{C_{5}^{2}}{10}\right)\frac{P_{6}}{A_{6}^{4}}$$.

1. $$\frac{A_{11}^{3}-A_{10}^{2}}{A_{9}^{1}}=\frac{\frac{11!}{(11-3)!}-\frac{10!}{(10-2)!}}{\frac{9!}{(9-1)!}}=\frac{\frac{11!}{8!}-\frac{10!}{8!}}{\frac{9!}{8!}}=\frac{11!-10!}{9!}$$

   $$\frac{11\cdot 10! — 10!}{9!}=\frac{10\cdot 10!}{9!}=\frac{10\cdot 10\cdot 9!}{9!}=100$$

2. $$\frac{A_{12}^{4}\cdot A_{7}^{7}}{A_{11}^{9}}=\frac{\frac{12!}{(12-4)!}\cdot \frac{7!}{(7-7)!}}{\frac{11!}{(11-9)!}}=\frac{\frac{12!}{8!}\cdot 7!}{\frac{11!}{2!}}$$

   $$\frac{12\cdot 11\cdot 10\cdot 9\cdot 7! \cdot 2!}{8!\cdot 11!}=\frac{12}{8}\cdot 2=3$$

3. $$\frac{A_{6}^{3}}{P_{4}}+\frac{A_{11}^{6}}{11P_{6}}=\frac{\frac{6!}{(6-3)!}}{4!}+\frac{\frac{11!}{(11-6)!}}{11\cdot 6!}$$

   $$=\frac{6!}{3!\cdot 4!}+\frac{11!}{5!\cdot 11\cdot 6!}=\frac{6\cdot 5\cdot 4!}{3!\cdot 4!}+\frac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6!}{5!\cdot 11\cdot 6!}$$

   $$=5+42=47$$

4. $$\left(\frac{C_{11}^{7}}{10}-\frac{C_{5}^{2}}{10}\right)\frac{P_{6}}{A_{6}^{4}}=\left(\frac{\frac{11!}{7!\,4!}}{10}-\frac{\frac{5!}{2!\,3!}}{10}\right)\cdot \frac{6!}{\frac{6!}{2!}}$$

   $$=\left(\frac{11\cdot 10\cdot 9\cdot 8}{4!\cdot 10}-\frac{5\cdot 4}{2\cdot 10}\right)\cdot 2=\left(\frac{11\cdot 9\cdot 8}{4\cdot 3\cdot 2}-1\right)\cdot 2$$

   $$=(33-1)\cdot 2=64$$

Ответ

1) 100; 2) 3; 3) 47; 4) 64.