Решебник по алгебре 9 класс Макарычев Задание 286

 

\[\boxed{\text{286}\text{\ (286)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 13 \cdot (5x - 1) -\]

\[- 15 \cdot (4x + 2) < 0\]

\[65x - 13 - 60x - 30 < 0\]

\[5x - 43 < 0\ \]

\[x < 8,6.\]

\[\textbf{б)}\ 6 \cdot (7 - 0,2x) -\]

\[- 5 \cdot (8 - 0,4x) > 0\]

\[42 - 1,2x - 40 + 2x > 0\]

\[0,8x + 2 > 0\]

\[0,8x > - 2\]

\[x > - 2,5.\]

\[\boxed{\text{286.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

Пояснение.

Решение.

\[\textbf{а)}\ (x + 25)(x - 30) < 0\]

\[x \in ( - 25;30).\]

\[\textbf{б)}\ (x + 6)(x - 6) > 0\]

\[x \in ( - \infty;\ - 6) \cup (6; + \infty).\]

\[\textbf{в)}\ \left( x - \frac{1}{3} \right)\left( x - \frac{1}{5} \right) \leq 0\]

\[(x - 0,2)\left( x - \frac{1}{3} \right) \leq 0\]

\[x \in \left\lbrack 0,2;\frac{1}{3} \right\rbrack.\]

\[\textbf{г)}\ (x + 0,1)(x + 6,3) \geq 0\]

\[(x + 6,3)(x + 0,1) \geq 0\]

\[x \in ( - \infty;\ - 6,3\rbrack \cup \lbrack - 0,1; + \infty).\]