ГДЗ по алгебре 9 класс Макарычев Задание 74

\[\boxed{\text{74\ (74).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 3 \cdot (x + 4)^{2} = 10x + 32\]

\[3 \cdot \left( x^{2} + 8x + 16 \right) = 10x + 32\]

\[3x^{2} + 24x - 10x + 48 - 32 = 0\]

\[3x^{2} + 14x + 16 = 0\]

\[D_{1} = 7^{2} - 3 \cdot 16 = 49 - 48 = 1\]

\[x_{1} = \frac{- 7 + 1}{3} = - 2;\]

\[x_{2} = \frac{- 7 - 1}{3} = - \frac{8}{3} = - 2\frac{2}{3}.\]

\[Ответ:x = - 2;\ \ x = - 2\frac{2}{3}.\]

\[\textbf{б)}\ 31x + 77 = 15 \cdot (x + 1)^{2}\]

\[31x + 77 = 15 \cdot \left( x^{2} + 2x + 1 \right)\]

\[31x + 77 = 15x^{2} + 30x + 15\]

\[15x^{2} + 30x - 31x + 15 - 77 = 0\]

\[15x^{2} - x - 62 = 0\]

\[D = 1 + 4 \cdot 15 \cdot 62 = 3721 = 61^{2}\]

\[x_{1} = \frac{1 - 61}{30} = - \frac{60}{30} = - 2;\]

\[x_{2} = \frac{62}{30} = 2\frac{2}{30} = 2\frac{1}{15}.\]

\[Ответ:x = - 2;\ \ x = 2\frac{1}{15}.\]