Решебник по алгебре 9 класс Макарычев Задание 333

 

\[\boxed{\text{333\ (333).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ (2x + 5)(x - 17) \geq 0\]

\[2 \cdot (x + 2,5)(x - 17) \geq 0\]

\[x \in ( - \infty; - 2,5\rbrack \cup \lbrack 17; + \infty).\]

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

\[2x(x + 9)(x - 4) \geq 0\]

\[x \in \lbrack - 9;0\rbrack \cup \lbrack 4;\ + \infty).\]

\[\boxed{\text{333.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{x^{2} + 1}{x} + \frac{x}{x^{2} + 1} = 2\frac{1}{2}\]

\[Пусть\ \ \ \ \ t = \frac{x^{2} + 1}{x},\ тогда\ \ \ \ \]

\[t + \frac{1}{t} = 2\frac{1}{2},\ \]

\[\frac{t^{2} + 1}{t} = \frac{5}{2},\ \ 2t^{2} + 2 = 5t,\ \]

\[2t^{2} - 5t + 2 = 0\]

\[D = 25 - 4 \cdot 2 \cdot 2 = 9\]

\[t_{1,2} = \frac{5 \pm 3}{4} = 2;\frac{1}{2};\ \]

\[1)\ \frac{x^{2} + 1}{x} = 2,\]

\[\text{\ \ }x^{2} - 2x + 1 = 0,\ \ \]

\[(x - 1)^{2} = 0,\ \ x_{1} = 1;\]

\[2)\ \frac{x^{2} + 1}{x} = \frac{1}{2},\]

\[\ \ 2x² - x + 2 = 0\]

\[D = 1 - 4 \cdot 2 \cdot 2 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[\textbf{б)}\ \frac{x^{2} + 2}{3x - 2} + \frac{3x - 2}{x^{2} + 2} = 2\frac{1}{6}\]

\[Пусть\ \ t = \frac{x^{2} + 2}{3x - 2},\ \ тогда\ \]

\[\ t + \frac{1}{t} = 2\frac{1}{6},\]

\[\frac{t^{2} + 1}{t} = \frac{13}{6},\]

\[\ \ 6t^{2} - 13t + 6 = 0,\]

\[D = 13² - 4 \cdot 6 \cdot 6 = 25\]

\[t_{1,2} = \frac{13 \pm 5}{12} = \frac{3}{2};\frac{2}{3};\]

\[1)\ \frac{x^{2} + 2}{3x - 2} = \frac{3}{2},\]

\[\ \ 2x^{2} + 4 = 9x - 6,\]

\[\ \ 2x^{2} - 9x + 10 = 0,\]

\[D = 81 - 4 \cdot 2 \cdot 10 = 1\]

\[x_{1,2} = \frac{9 \pm 1}{4},\ \ x_{1} = 2,\]

\[\text{\ \ }x_{2} = 2,5;\]

\[2)\ \frac{x^{2} + 2}{3x - 2} = \frac{2}{3},\]

\[\ \ 3x^{2} + 6 = 6x - 4,\]

\[\ \ 3x^{2} - 6x + 10 = 0\]

\[D = 9 - 3 \cdot 10 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:\ \ \ а)\ 1;\ \ \ \ б)\ 2;2,5.\]