Решебник по алгебре 9 класс Макарычев Задание 285

 

\[\boxed{\text{285}\text{\ (285)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 3x^{2} - 25x - 28 = 0\]

\[D = 625 + 4 \cdot 3 \cdot 28 =\]

\[= 961 = 31^{2}\]

\[x_{1} = \frac{25 + 31}{6} = \frac{28}{3};\ \ \ \ \]

\[x_{2} = \frac{25 - 31}{6} = - 1;\]

\[\Longrightarrow 3x^{2} - 25x - 28 =\]

\[= 3 \cdot \left( x - \frac{28}{3} \right)(x + 1) =\]

\[= (3x - 28)(x + 1).\]

\[\textbf{б)}\ 2x^{2} + 13x - 7 = 0\]

\[D = \ 169 + 4 \cdot 2 \cdot 7 = 225\]

\[x_{1} = \frac{- 13 + 15}{4} = \frac{1}{2};\ \ x_{2} =\]

\[= \frac{- 15 - 13}{4} = - 7;\]

\[\Longrightarrow 2x^{2} + 13x - 7 =\]

\[= 2 \cdot \left( x - \frac{1}{2} \right)(x + 7) =\]

\[= (2x - 1)(x + 7).\]

\[\boxed{\text{285.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

Пояснение.

Решение.

\[\textbf{а)}\ (x + 8)(x - 5) > 0\]

\[x \in ( - \infty;\ - 8) \cup (5; + \infty).\]

\[\textbf{б)}\ (x - 14)(x + 10) < 0\]

\[(x + 10)(x - 14) < 0\]

\[x \in ( - 10;14).\]

\[\textbf{в)}\ (x - 3,5)(x + 8,5) \geq 0\]

\[(x + 8,5)(x - 3,5) \geq 0\]

\[x \in ( - \infty;\ - 8,5\rbrack \cup \lbrack 3,5;\ + \infty).\]

\[\textbf{г)}\ \left( x + \frac{1}{3} \right)\left( x + \frac{1}{8} \right) \leq 0\]

\[x \in \left\lbrack - \frac{1}{3};\ - \frac{1}{8} \right\rbrack.\]