Решебник по алгебре 9 класс Макарычев Задание 647

 

\[\boxed{\text{647\ (646).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 2x² - 13x - 34 \geq 0\]

\[2x^{2} - 13x - 34 = 0\]

\[D = 169 + 4 \cdot 2 \cdot 34 = 441\]

\[x_{1,2} = \frac{13 \pm 21}{4} = - 2;\ 8,5;\]

\[\Longrightarrow 2 \cdot (x + 2)(x - 8,5) \geq 0.\]

\[\]

\[x \in ( - \infty; - 2\rbrack \cup \lbrack 8,5; + \infty).\]

\[\textbf{б)}\ 10x - 4x^{2} < 0\]

\[4x^{2} - 10x > 0\]

\[x(2x - 5) > 0\]

\[x(x - 2,5) > 0\]

\[\]

\[x \in ( - \infty;0) \cup (2,5;\ + \infty).\]

\[\textbf{в)}\ \frac{x - 4}{2x + 5} \leq 0\]

\[(x - 4)(2x + 5) \leq 0,\ \ x \neq - 2,5\]

\[\]

\[x \in ( - 2,5;4\rbrack.\]

\[\boxed{\text{647.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ a_{1} = - 10\frac{1}{2},\ \ a_{2} = - 10\frac{1}{4},\ \ \]

\[d = a_{2} - a_{1} = - 10\frac{1}{4} + 10\frac{1}{2} =\]

\[= - \frac{41}{4} + \frac{21}{2} = \frac{1}{4},\]

\[a_{n} = a_{1} + d(n - 1) > 0 \Longrightarrow\]

\[- 10\frac{1}{2} + \frac{1}{4}n - \frac{1}{4} > 0\]

\[\frac{1}{4}n > 10\frac{3}{4}\]

\[n > 43 \Longrightarrow n = 44,\]

\[a_{44} = - 10\frac{1}{2} + \frac{1}{4} \cdot (44 - 1) =\]

\[= - \frac{42}{4} + \frac{43}{4} = \frac{1}{4}.\]

\[\textbf{б)}\ a_{1} = 8\frac{1}{2},\ \ a_{2} = 8\frac{1}{3},\ \]

\[d = a_{2} - a_{1} = 8\frac{1}{3} - 8\frac{1}{2} =\]

\[= \frac{25}{3} - \frac{17}{2} = - \frac{1}{6},\]

\[a_{n} = \frac{25}{3} - \frac{1}{6} \cdot (n - 1) \Longrightarrow\]

\[\frac{25}{3} - \frac{n}{6} + \frac{1}{6} < 0\]

\[n > 50 + 1 \Longrightarrow n > 51 \Longrightarrow\]

\[\Longrightarrow n = 52,\]

\[a_{52} = 8,5 - \frac{1}{6} \cdot (53 - 1) = - \frac{1}{6}.\]