Решебник по алгебре 9 класс Мерзляк Задание 476

\[\boxed{\text{476\ (476).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\left\{ \begin{matrix} 2 \cdot (x - 3) \geq - 3 \cdot (x + 2) \\ \frac{7x}{3} \leq 1 - \frac{x}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \cdot 6\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} 2x - 6 \geq - 3x - 6 \\ \frac{14x + 3x}{6} \leq 1\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 5x \geq 0\ \ \\ \frac{17x}{6} \leq 1 \\ \end{matrix}\ \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x \geq 0 \\ x \leq \frac{6}{17} \\ \end{matrix} \right.\ \]

\[Ответ:x \in \left\lbrack 0;\frac{6}{17} \right\rbrack.\]

\[\boxed{\text{476.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ x^{2} - 5x - 10 = 0\]

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

\[x_{1} \cdot x_{2} = - \frac{10}{1} = - 10\]

\[2)\ 2x^{2} + 6x - 7 = 0\]

\[x_{1} + x_{2} = - \frac{6}{2} = - 3\]

\[x_{1} \cdot x_{2} = - \frac{7}{2} = - 3,5\]

\[3) - \frac{1}{3}x^{2} + 8x - 1 = 0\]

\[x_{1} + x_{2} = 8\ :\frac{1}{3} = 8 \cdot 3 = 24\]

\[x_{1} \cdot x_{2} = 1\ :\frac{1}{3} = 3\]