ГДЗ по алгебре 8 класс Макарычев Задание 691

\[\boxed{\text{691\ (691).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[График\ функции\ пересекается\]

\[\ с\ \text{OX}\ при\ y = 0.\]

\[\textbf{а)}\frac{2x - 5}{x + 3} = 0\]

\[Ответ:в\ точке\ \ (2,5;0).\]

\[\textbf{б)}\frac{(x - 4)(3x - 15)}{x - 9} = 0\]

\[\left\{ \begin{matrix} x - 4 = 0 \\ 3x - 15 = 0 \\ x - 9 \neq 0\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 4 \\ x = 5 \\ x \neq 9 \\ \end{matrix} \right.\ \]

\[Ответ:в\ точках\ (4;0)\ и\ (5;0).\]

\[\textbf{в)}\frac{x^{2} - 5x + 6}{x - 2} = 0\]

\[\left\{ \begin{matrix} x^{2} - 5x + 6 = 0* \\ x - 2 \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ \]

\[*D = 25 - 24 = 1\]

\[x_{1,2} = \frac{5 \pm 1}{2} = 3;2\]

\[\left\{ \begin{matrix} x = 3 \\ x = 2 \\ x \neq 2 \\ \end{matrix} \right.\ \]

\[Ответ:в\ точке\ (3;0).\]

\[\textbf{г)}\frac{x^{3} - 7x^{2} + 12x}{x - 3} = 0\]

\[\frac{x \cdot \left( x^{2} - 7x + 12 \right)}{x - 3} = 0\]

\[\left\{ \begin{matrix} x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 7x + 12 = 0* \\ x - 3 \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[*D = 49 - 48 = 1\]

\[x_{1,2} = \frac{7 \pm 1}{2} = 4;3\]

\[\left\{ \begin{matrix} x = 0 \\ x = 4 \\ x = 3 \\ x \neq 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[Ответ:в\ точках\ (0;0)\ и\ (4;0)\text{.\ }\]