ГДЗ по алгебре 9 класс Макарычев Задание 339

 

\[\boxed{\text{339\ (339).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = ax + b \Longrightarrow общий\ вид\ \]

\[уравнения\ прямой.\]

\[\textbf{а)}\ (0;0)\ и\ (0,6;\ - 2,4) \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 0 = 0 \cdot a + b\ \ \ \ \ \ \\ - 2,4 = 0,6a + b \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} b = 0\ \ \ \\ a = - 4 \\ \end{matrix} \right.\ \Longrightarrow y = - 4x.\]

\[\textbf{б)}\ (0;4)\ и\ \ ( - 2,5;0) \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 4 = 0 \cdot a + b\ \ \ \\ 0 = - 2,5a + b \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} b = 4\ \ \ \ \\ a = 1,6 \\ \end{matrix} \right.\ \Longrightarrow y = 1,6x + 4.\]

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

\[\textbf{а)}\ x² - 5x - 50 < 0,\]

\[\ \ по\ теореме\ Виета:\]

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

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

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

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

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

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

\[m_{2} = 1.\]

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

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

\[\textbf{в)}\ 3y² + 4y - 4 > 0\]

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

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

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

\[\textbf{г)}\ 8p² + 2p \geq 21\]

\[8p² + 2p - 21 \geq 0\]

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

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

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

\[\textbf{д)}\ 12x - 9 \leq 4x²\]

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

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

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

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

\[\textbf{е)} - 9x^{2} < 1 - 6x\]

\[9x² - 6x + 1 > 0\ \]

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

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