Решебник по алгебре 9 класс Макарычев Задание 240

 

\[\boxed{\text{240}\text{\ (240)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Прямая\ является\ осью\ \]

\[симметрии\ параболы,\]

\[\ когда\ на\ этой\ \]

\[прямой\ лежит\ вершина\ \]

\[параболы.\]

\[y = ax^{2} - 16x + 1\]

\[x_{b} = - \frac{b}{2a} = - \frac{- 16}{2a} = \frac{8}{a}\]

\[x = 4:\]

\[4 = \frac{8}{a}\]

\[a = 2.\]

\[Ответ:при\ a = 2.\]

\[\boxed{\text{240.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x² + x - 9 = \frac{9}{x}\]

\[x^{3} + x^{2} - 9x - 9 = 0,\ \ x \neq 0\]

\[x^{2}(x + 1) - 9 \cdot (x + 1) = 0\]

\[\left( x^{2} - 9 \right)(x + 1) = 0\]

\[(x - 3)(x + 3)(x + 1) = 0\]

\[x_{1} = 3,\ \ x_{2} = - 3,\ \ \]

\[x_{3} = - 1;\]

\[y(3) = \frac{9}{3} = 3 \Longrightarrow (3;3);\]

\[y( - 3) = \frac{9}{- 3} = - 3 \Longrightarrow ( - 3;\ - 3);\]

\[y( - 1) = \frac{9}{- 1} = - 9 \Longrightarrow ( - 1;\ - 9).\]

\[\textbf{б)}\ x² + 6x - 4 = \frac{24}{x}\]

\[x^{3} + 6x^{2} - 4x - 24 = 0,\]

\[\ \ x \neq 0\]

\[x^{2}(x + 6) - 4 \cdot (x + 6) = 0\]

\[\left( x^{2} - 4 \right)(x + 6) = 0\]

\[(x - 2)(x + 2)(x + 6) = 0\]

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

\[\text{\ \ }x_{3} = - 6;\]

\[y(2) = \frac{24}{2} = 12 \Longrightarrow (2;12);\]

\[y( - 2) = \frac{24}{- 2} = - 12 \Longrightarrow\]

\[\Longrightarrow ( - 2;\ - 12)\mathbf{;}\]

\[y( - 6) = \frac{24}{- 6} = - 4 \Longrightarrow\]

\[\Longrightarrow ( - 6; - 4).\]