ГДЗ по алгебре 9 класс Макарычев Задание 412

\[\boxed{\text{412\ (412).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 25x^{2} + 6x \leq 0\]

\[x(25x + 6) \leq 0\]

\[25x\left( x + \frac{6}{25} \right) \leq 0\]

\[x \in \lbrack - 0,24;0\rbrack\text{.\ }\]

\[\textbf{б)}\ x^{2} - 169 > 0\]

\[(x + 13)(x - 13) > 0\]

\[x \in ( - \infty;\ - 13) \cup (13;\ + \infty).\]

\[\textbf{в)}\ 4x^{2} - 225 \leq 0\]

\[(2x - 15)(2x + 15) \leq 0\]

\[4 \cdot (x + 7,5)(x - 7,5) \leq 0\]

\[x \in \lbrack - 7,5;7,5\rbrack.\]

\[\textbf{г)}\ y^{2} < 10y + 24\]

\[y^{2} - 10y - 24 < 0\]

\[D_{1} = 25 + 24 = 49\]

\[y_{1} = 5 + 7 = 12;\ \ y_{2} =\]

\[= 5 - 7 = - 2.\]

\[(y + 2)(y - 12) < 0\]

\[y \in ( - 2;12).\]

\[\textbf{д)}\ 15y^{2} + 30 > 22y + 7\]

\[15y^{2} - 22y + 23 > 0\]

\[D = 11^{2} - 15 \cdot 23 =\]

\[= 121 - 345 < 0 \Longrightarrow\]

\[\Longrightarrow y - любое\ число.\]

\[y \in ( - \infty; + \infty).\]

\[\textbf{е)}\ 3y^{2} - 7 \leq 26y + 70\]

\[3y^{2} - 26y - 77 \leq 0\]

\[D = 13^{2} + 3 \cdot 77 = 400\]

\[y_{1,2} = \frac{13 \pm 20}{3} = - \frac{7}{3};\ \ 11;\]

\[3 \cdot \left( y + \frac{7}{3} \right)(y - 11) \leq 0\]

\[y \in \left\lbrack - 2\frac{1}{3};11 \right\rbrack.\]