ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 338

\[\boxed{\text{339.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x² - 5x - 50 < 0,\]

\[\ \ по\ теореме\ Виета:\]

\[x_{1} = 10,\ x_{2} = - 5,\]

\[\Longrightarrow (x - 10)(x + 5) < 0 \Longrightarrow\]

\[\Longrightarrow x \in ( - 5;10).\]

\[\textbf{б)} - m^{2} - 8x + 9 \geq 0,\]

\[\text{\ \ }m^{2} + 8m - 9 \leq 0,\]

\[по\ теореме\ Виета:\ \ m_{1} = - 9,\ \ \]

\[m_{2} = 1.\]

\[(m + 9)(m - 1) \leq 0 \Longrightarrow\]

\[\Longrightarrow m \in \lbrack - 9;1\rbrack.\]

\[\textbf{в)}\ 3y² + 4y - 4 > 0\]

\[D = 4 + 3 \cdot 4 = 16\]

\[y_{1,2} = \frac{- 2 \pm 4}{3} = - 2;\ \frac{2}{3};\ \]

\[y \in ( - \infty;\ - 2) \cup \left( \frac{2}{3};\ + \infty \right).\]

\[\textbf{г)}\ 8p² + 2p \geq 21\]

\[8p² + 2p - 21 \geq 0\]

\[D = 1 + 8 \cdot 21 = 169\]

\[p_{1,2} = \frac{- 1 \pm 13}{8} = 1,5;\ - 1,75.\]

\[p \in ( - \infty; - 1,75\rbrack \cup \lbrack 1,5;\ + \infty).\]

\[\textbf{д)}\ 12x - 9 \leq 4x²\]

\[4x^{2} - 12x + 9 \geq 0\]

\[D = 36 - 4 \cdot 9 = 0\]

\[(2x - 3)^{2} \geq 0\]

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

\[\textbf{е)} - 9x^{2} < 1 - 6x\]

\[9x² - 6x + 1 > 0\ \]

\[(3x - 1)^{2} > 0\]

\[x \in \left( - \infty;\frac{1}{3} \right) \cup \left( \frac{1}{3}; + \infty \right).\]