Решебник по алгебре 9 класс Макарычев Задание 454

 

\[\boxed{\text{454\ (454).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[D_{1} = 1 + 168 = 169\]

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

\[(x + 14)(x - 12) < 0\]

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

\[\textbf{б)}\ 5x^{2} + x - 2 < 0\]

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

\[x_{1,2} = \frac{- 1 \pm 11}{30};\ \ x_{1,2} = - \frac{2}{5};\frac{1}{3}\text{.\ }\]

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

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

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

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

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

\[2 \cdot (x + 14)(x - 1,5) > 0\]

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

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

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

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

\[5 \cdot (x + 25)(x - 1,2) < 0\]

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

\(\boxed{\text{454}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)

\[\textbf{а)}\ x² + y² - 6x - 4y + 13 \leq 0\]

\[\left( x^{2} - 6x + 9 \right) + \left( y^{2} - 4y + 4 \right) \leq 0\]

\[(x - 3)^{2} + (y - 2)^{2} \leq 0,\]

\[точка\ (3;2).\]

\[\textbf{б)}\ x² - 4x - y + 5 \geq 0\]

\[y \leq x^{2} - 4x + 4 + 1\]

\[y \leq (x - 2)^{2} + 1.\]

\[Множество\ точек,\ \]

\[принадлежащих\ параболе\ \]

\[y = (x - 2)^{2} + 1\ и\ точек,\ \]

\[расположенных\ ниже.\]