ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 946

\[\boxed{\text{946.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ 0,2x² - 0,2 \cdot (x - 6) \cdot\]

\[\cdot (x + 6) > 3,6x\]

\[0,2x^{2} - 0,2x^{2} - 1,2x + 1,2x +\]

\[+ 7,2 > 3,6x\]

\[7,2 > 3,6x\]

\[3,6x < 7,2\]

\[x < 2 \Longrightarrow \ \ x \in ( - \infty;2)\]

\[\textbf{б)}\ (2x - 5)^{2} - 0,5x < (2x - 1) \cdot\]

\[\cdot (2x + 1) - 15\]

\[4x^{2} - 20x + 25 - 0,5x < 4x^{2} -\]

\[- 1 - 15\]

\[- 20x - 0,5x < - 16 - 25\]

\[- 20,5x < - 41\]

\[x > 2 \Longrightarrow \ \ x \in (2;\ + \infty)\]

\[\textbf{в)}\ (12x - 1)(3x + 1) < 1 +\]

\[+ (6x + 2)²\]

\[36x^{2} + 12x - 3x - 1 < 1 +\]

\[+ 36x^{2} + 24x + 4\]

\[9x - 24x < 5 + 1\]

\[- 15x < 6\]

\[x > - 0,4 \Longrightarrow \ \ x \in ( - 0,4;\ + \infty)\]

\[\textbf{г)}\ (4y - 1)^{2} > (2y + 3)(8y - 1)\]

\[16y^{2} - 8y + 1 > 16y^{2} - 2y +\]

\[+ 24y - 3\]

\[- 30y > - 4\]

\[y < \frac{4}{30}\]

\[\ y < \frac{2}{15} \Longrightarrow \ \ y \in \left( - \infty;\frac{2}{15} \right)\]