Решебник по алгебре 9 класс Макарычев Задание 907

 

\[\boxed{\text{907\ (907).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x² - x - 42 = (x - 7)(x + 6)\]

\[x^{2} - x - 42 = 0\]

\[D = 1 + 168 = 169\]

\[x_{1} = \frac{1 - 13}{2} = - 6,\ \ \]

\[x_{2} = \frac{1 + 13}{7} = 7.\]

\[\textbf{б)}\ y² + 9y + 18 = (y + 3)(y + 6)\]

\[y^{2} + 9y + 18 = 0\]

\[D = 81 - 72 = 9\]

\[y_{1} = \frac{- 9 + 3}{2} = - 3,\ \ \]

\[y_{2} = \frac{- 9 - 3}{2} = - 6.\]

\[\textbf{в)}\ 81x² + 18x + 1 = (9x + 1)^{2} =\]

\[= (9x + 1)(9x + 1)\]

\[\textbf{г)}\ 16b² - 24b + 9 = (4b - 3)^{2} =\]

\[= (4b - 3)(4b - 3)\]

\[\textbf{д)}\ 6x² - x - 1 =\]

\[= 6 \cdot \left( x - \frac{1}{2} \right)\left( x + \frac{1}{3} \right) =\]

\[= (2x - 1)(3x + 1)\]

\[6x² - x - 1 = 0\]

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{1 + 5}{12} = \frac{6}{12} = \frac{1}{2},\ \ \]

\[x_{2} = \frac{1 - 5}{12} = - \frac{4}{12} = - \frac{1}{3}.\]

\[\textbf{е)}\ 3a² - 13a - 10 =\]

\[= 3 \cdot (a - 5)\left( a + \frac{2}{3} \right) =\]

\[= (a - 5)(3a + 2)\]

\[3a^{2} - 13a - 10 = 0\]

\[D = 169 + 120 = 289\]

\[a_{1} = \frac{13 + 17}{6} = 5,\ \ \]

\[a_{2} = \frac{13 - 17}{6} = - \frac{4}{6} = - \frac{2}{3}.\]