ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 1039

\[\boxed{\text{1039.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ 0,01 \cdot (1 - 3x) > 0,02x + 3,01\]

\[0,01 - 0,03x > 0,02x + 3,01\]

\[- 0,03x - 0,02x > 3,01 - 0,01\]

\[- 0,05x > 3\ \ \ \ \ \ |\ :( - 0,05)\]

\[x < - 60 \Longrightarrow \ \ x \in ( - \infty;\ - 60)\]

\[\textbf{б)}\ 12 \cdot (1 - 12x) +\]

\[+ 100x > 36 - 49x\]

\[12 - 144x + 100x > 36 - 49x\]

\[- 44x + 49x > 36 - 12\]

\[5x > 24\ \ \ |\ :5\]

\[x > 4,8 \Longrightarrow \ \ \ \ \ \ \ x \in (4,8;\ + \infty)\]

\[\textbf{в)}\ (0,6y - 1) - 0,2 \cdot\]

\[\cdot (3y + 1) < 5y - 4\]

\[0,6y - 1 - 0,6y - 0,2 < 5y - 4\]

\[- 1,2 + 4 < 5y\]

\[2,8 < 5y\ \ \ \ \ \ |\ :5\]

\[y > \frac{2,8}{5}\]

\[y > 0,56 \Longrightarrow \ \ \ \ \ \ \ \ y \in (0,56;\ + \infty)\]

\[\textbf{г)}\frac{2}{3} \cdot (6x + 4) - \frac{1}{6} \cdot\]

\[\cdot (12x - 5) \leq 4 - 6x\ \ \ \ \ | \cdot 6\]

\[4 \cdot (6x + 4) -\]

\[- (12x - 5) \leq 24 - 36x\]

\[24x + 16 - 12x + 5 \leq 24 - 36x\]

\[12x + 36x \leq 24 - 21\]

\[48x \leq 3\ \ \ \ |\ :48\]

\[x \leq \frac{1}{16} \Longrightarrow \text{\ \ x} \in \left( - \infty;\frac{1}{16} \right\rbrack\]

\[\textbf{д)}\ (3a + 1)(a - 1) -\]

\[- 3a^{2} > 6a + 7\]

\[3a^{2} - 3a + a - 1 -\]

\[- 3a^{2} > 6a + 7\]

\[- 2a - 6a > 7 + 1\]

\[- 8a > 8\ \ \ |\ :( - 8)\]

\[a < - 1 \Longrightarrow \ \ a \in ( - \infty;\ - 1)\ \]

\[\textbf{е)}\ 15x² - (5x - 2)(3x + 1) <\]

\[< 7x - 8\]

\[15x^{2} - 15x^{2} - 5x + 6x + 2 -\]

\[- 7x + 8 < 0\]

\[- 6x < - 10\ \ \ \ \ |\ :( - 6)\]

\[x > \frac{10}{6}\]

\[x > \frac{5}{3} \Longrightarrow \ \ x \in \left( 1\frac{2}{3};\ + \infty \right)\]