Решебник по алгебре 9 класс Макарычев Задание 327

 

\[\boxed{\text{327\ (327).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ (x - 2)(x - 5)(x - 12) > 0\]

\[x \in (2;5) \cup (12; + \infty).\]

\[\textbf{б)}\ (x + 7)(x + 1)(x - 4) < 0\]

\[x \in ( - \infty;\ - 7) \cup ( - 1;4).\]

\[\textbf{в)}\ x(x + 1)(x + 5)(x - 8) > 0\]

\[(x + 5)(x + 1)x(x - 8) > 0\]

\[x \in ( - \infty;\ - 5) \cup ( - 1;0) \cup\]

\[\cup (8;\ + \infty).\]

\[\boxed{\text{327.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{3y^{3} + 12y^{2} - 27y - 108}{y^{2} - 16} = 0;\]

\[ОДЗ:\ \ \ y^{2} - 16 \neq 0,\ \ y^{2} \neq 16,\ \ \]

\[y \neq \pm 4.\]

\[3y^{3} + 12y^{2} - 27y - 108 = 0|\ :3\]

\[y^{3} + 4y^{2} - 9y - 36 = 0\]

\[y^{2}(y + 4) - 9 \cdot (y + 4) = 0\]

\[\left( y^{2} - 9 \right)(y + 4) = 0\]

\[1)\ y^{2} - 9 = 0\]

\[y^{2} = 9\]

\[y = \pm 3.\]

\[2)\ y + 4 = 0\]

\[y = - 4 \Longrightarrow не\ удовлетворяет\ \]

\[ОДЗ.\]

\[\textbf{б)}\ \frac{y^{3} + 6y^{2} - y - 6}{y^{3} - 36y} = 0;\]

\[ОДЗ:\ \ y^{3} - 36y \neq 0;\ \ \ \ \]

\[y\left( y^{2} - 36 \right) \neq 0;\]

\[y(y - 6)(y + 6) \neq 0;\ \ \]

\[y \neq 0;\ y \neq \pm \ 6.\]

\[y^{3} + 6y^{2} - y - 6 = 0\]

\[y^{2}(y + 6) - (y + 6) = 0\]

\[\left( y^{2} - 1 \right)(y + 6) = 0\]

\[(y - 1)(y + 1)(y + 6) = 0\]

\[y_{1} = 1;\ \ y_{2} = - 1;\ \ y_{3} = - 3 \Longrightarrow\]

\[\Longrightarrow не\ удовлетворяет\ ОДЗ.\]

\[Ответ:а) - 3;3;\ \ \ б) - 1;1.\]