ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 718

\[\boxed{\text{718.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)} - x^{2} - 2x + 168 > 0\]

\[x^{2} + 2x - 168 < 0\]

\[D = 1 + 168 = 169\]

\[x_{1,2} = - 1 \pm 13 = - 14;12;\]

\[(x + 4)(x - 12) < 0\]

\[x \in ( - 14;12).\]

\[Ответ:при\ x \in ( - 14;12).\]

\[\textbf{б)}\ 5x² + x - 2 < 0\]

\[D = 1 + 4 \cdot 15 \cdot 2 = 121\]

\[x_{1} = \frac{- 1 - 11}{30} = - \frac{12}{30} = - \frac{2}{5};\]

\[x_{2} = \frac{- 1 + 11}{30} = \frac{1}{3}.\]

\[15 \cdot \left( x + \frac{2}{5} \right)\left( x - \frac{1}{3} \right) < 0\]

\[x \in \left( - \frac{2}{5};\frac{1}{3} \right).\]

\[Ответ:\ при\ x \in \left( - \frac{2}{5};\frac{1}{3} \right)\text{.\ }\]

\[\textbf{в)}\ \frac{x + 14}{3 - 2x} < 0\]

\[(x + 14)(3 - 2x) < 0\]

\[(x + 14)(2x - 3) > 0\]

\[x \in ( - \infty;\ - 14) \cup (1,5;\ + \infty).\]

\[Ответ:при\ \]

\[x \in ( - \infty;\ - 14) \cup (1,5;\ + \infty).\]

\[\textbf{г)}\ \frac{6 - 5x}{x + 25} > 0\ \]

\[(6 - 5x)(x + 25) > 0\]

\[(5x - 6)(x + 25) < 0\]

\[(x - 1,2)(x + 25) < 0\]

\[x \in ( - 25;1,2).\]

\[Ответ:при\ x \in ( - 25;1,2).\]