Решебник по алгебре 9 класс Макарычев Задание 749

 

\[\boxed{\text{749\ (749).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\frac{12!}{9!} = \frac{9! \cdot 10 \cdot 11 \cdot 12}{9!} = 1320;\]

\[\textbf{б)}\frac{14!}{12!} = \frac{12! \cdot 13 \cdot 14}{12!} = 182;\]

\[\textbf{в)}\frac{30!}{29! \cdot 2!} = \frac{29! \cdot 30}{29! \cdot 2!} = 15;\]

\[\textbf{г)}\ \frac{36!}{2! \cdot 34!} = \frac{34! \cdot 35 \cdot 36}{2! \cdot 34!} = 630;\]

\[\textbf{д)}\ \frac{15!}{2! \cdot 16!} = \frac{15!}{2! \cdot 15! \cdot 16} = \frac{1}{32};\]

\[\textbf{е)}\ \frac{25!}{23! \cdot 5!} = \frac{23! \cdot 24 \cdot 25}{23! \cdot 5!} =\]

\[= \frac{2 \cdot 3 \cdot 4 \cdot 5 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} = 5.\]

\[\boxed{\text{749.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 2 \cdot (5x - 1)^{2} + 35x - 11 = 0\]

\[2 \cdot (5x - 1)^{2} + 7 \cdot (5x - 1) - 4 =\]

\[= 0\]

\[Пусть\ c = (5x - 1):\]

\[2c^{2} + 7c - 4 = 0\]

\[D = 49 + 32 = 81\]

\[c_{1} = \frac{- 7 + 9}{4} = \frac{1}{2},\ \ \]

\[c_{2} = \frac{- 7 - 9}{4} = - 4,\]

\[1)\ 5x - 1 = \frac{1}{2}\]

\[5x = \frac{3}{2}\]

\[x = \frac{3}{5 \cdot 2} = 0,3;\]

\[2)\ 5x - 1 = - 4\]

\[5x = - 3\]

\[x = - 0,6.\]

\[Ответ:x = 0,3;\ x = - 0,6.\]

\[Пусть\ c = x^{2} + x - 3:\]

\[c^{2} + 12c + 27 = 0\]

\[D = 144 - 108 = 36,\]

\[c_{1} = \frac{- 12 + 6}{2} = - 3,\ \ \]

\[c_{2} = \frac{- 12 - 6}{2} = - 9,\]

\[1)\ x² + x - 3 = - 3\]

\[x^{2} + x = 0\]

\[x(x + 1) = 0\]

\[x_{1} = 0,\ \ x_{2} = - 1,\]

\[2)\ x² + x - 3 = - 9\]

\[x^{2} + x + 6 = 0\]

\[D = 1 - 24 < 0 \Longrightarrow корней\ нет.\]

\[Ответ:x = 0;\ x = - 1.\]