ГДЗ по алгебре 9 класс Макарычев Задание 952

\[\boxed{\text{952\ (952).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[= 0\]

\[Пусть\ c = (5x - 1):\]

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

\[D = 49 + 32 = 81\]

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

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

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

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

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

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

\[5x = - 3\]

\[x = - 0,6.\]

\[Ответ:x = 0,3;\ x = - 0,6.\]

\[Пусть\ c = x^{2} + x - 3:\]

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

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

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

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

\[1)\ x² + x - 3 = - 3\]

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

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

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

\[2)\ x² + x - 3 = - 9\]

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

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

\[Ответ:x = 0;\ x = - 1.\]