ГДЗ по алгебре 9 класс Макарычев Задание 852

\[\boxed{\text{852\ (852).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{P_{6} - P_{4}}{P_{5}} = \frac{6! - 4!}{4! \cdot 5} =\]

\[= \frac{4!(5 \cdot 6 - 1)}{4! \cdot 5} = \frac{29}{5} = 5\frac{4}{5};\]

\[\textbf{б)}\ \frac{P_{12} + P_{13}}{P_{11}} = \frac{12! + 13!}{11!} =\]

\[= \frac{11! \cdot 12 + 11! \cdot 12 \cdot 13}{11!} =\]

\[= \frac{11! \cdot (12 + 12 \cdot 13)}{11!} =\]

\[= 12 + 12 \cdot 13 = 168;\]

\[\textbf{в)}\ \frac{A_{8}^{4} - A_{8}^{3}}{A_{7}^{3} - A_{7}^{2}} = \frac{\frac{8!}{4!} - \frac{8!}{5!}}{\frac{7!}{4!} - \frac{7!}{5!}} =\]

\[= \frac{\frac{8!}{4!} \cdot \ (1 - \frac{1}{5})}{\frac{7!}{4!} \cdot \left( 1 - \frac{1}{5} \right)} = 8;\]

\[\textbf{г)}\ \frac{C_{6}^{3} - C_{6}^{2}}{A_{6}^{2}} = \frac{\frac{6!}{3! \cdot 3!} - \frac{6!}{2! \cdot 4!}}{\frac{6!}{4!}} =\]

\[= \left( \frac{4 \cdot 5 \cdot 6}{2 \cdot 3} - \frac{5 \cdot 6}{2} \right) = \frac{30}{6} \cdot \frac{1}{30} =\]

\[= \frac{1}{6}.\]