Решебник по алгебре 9 класс Макарычев Задание 338

 

\[\boxed{\text{338\ (338).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{5x + 4}{x} < 4^{\backslash x}\]

\[\frac{5x + 4 - 4x}{x} < 0\]

\[\frac{x + 4}{x} < 0\]

\[x(x + 4) < 0\]

\[x \in ( - 4;0).\]

\[\textbf{б)}\ \frac{6x + 1}{x + 1} > 1^{\backslash x + 1}\]

\[\frac{6x + 1 - x - 1}{x + 1} > 0\]

\[\frac{5x}{x + 1} > 0\]

\[5x(x + 1) > 0\]

\[x \in ( - \infty;\ - 1) \cup (0; + \infty).\]

\[\textbf{в)}\ \frac{x}{x - 1} \geq 2^{\backslash x - 1}\]

\[\frac{x - 2x + 2}{x - 1} \geq 0\]

\[\frac{- x + 2}{x - 1} \geq 0\]

\[\frac{x - 2}{x - 1} \leq 0\]

\[(x - 2)(x - 1) \leq 0\]

\[x \in (1;2\rbrack.\]

\[\textbf{г)}\ \frac{3x - 1}{x + 2} \geq 1\]

\[\frac{3x - 1}{x + 2} - 1^{\backslash x + 2} \geq 0\]

\[\frac{3x - 1 - x - 2}{x + 2} \geq 0\]

\[\frac{2x - 3}{x + 2} \geq 0\]

\[(2x - 3)(x + 2) \geq 0\]

\[2 \cdot (x + 2)(x - 1,5) \geq 0\]

\[x \in ( - \infty;\ - 2) \cup \lbrack 1,5;\ + \infty).\]

\[\boxed{\text{338.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[3\frac{1}{4} \cdot \left( a + \frac{1}{a} \right) = a^{3} + \frac{1}{a^{3}}\]

\[\frac{13}{4} \cdot \left( a + \frac{1}{a} \right) = a^{3} + \frac{1}{a^{3}}\]

\[Пусть\ \ y = a + \frac{1}{a};\ \ \ \ \]

\[y^{3} = \left( a + \frac{1}{a} \right)^{3} =\]

\[= a^{3} + \frac{1}{a^{3}} + \frac{3}{a} + 3a;\]

\[a^{3} + \frac{1}{a^{3}} = y^{3} - 3 \cdot \left( a + \frac{1}{a} \right) =\]

\[= y^{3} - 3y:\ \]

\[\frac{13}{4}y = y^{3} - 3y\]

\[4y^{3} - 12y - 13y = 0\ \]

\[4y^{3} - 25y = 0\]

\[y \cdot \left( 4y^{2} - 25 \right) = 0\]

\[y(2y - 5)(2y + 5) = 0\]

\[4y(y - 2,5)(y + 2,5) = 0\]

\[y_{1} = 0;\ \ y_{2} = 2,5;\ \ y_{3} = - 2,5.\]

\[1)\ a + \frac{1}{a} = 0\]

\[a^{2} + 1 = 0 \Longrightarrow корней\ нет.\]

\[2)\ a + \frac{1}{a} = 2,5\]

\[a^{2} - 2,5a + 1 = 0\ \ \]

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

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

\[a_{1,2} = \frac{5 \pm 3}{4} = 2;\frac{1}{2}\text{.\ }\]

\[3)\ a + \frac{1}{a} = - 2,5\]

\[\left( a + \frac{1}{a} \right)\ в\ 3\frac{1}{4}\ больше \Longrightarrow\]

\[\Longrightarrow не\ подходят.\]

\[Ответ:a = \frac{1}{2};\ \ a = 2.\]