Решебник по алгебре 9 класс Макарычев Задание 337

 

\[\boxed{\text{337\ (337).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{x - 8}{x + 4} > 2\]

\[\frac{x - 8}{x + 4} - 2^{\backslash x + 4} > 0\]

\[\frac{x - 8 - 2x - 8}{x + 4} > 0\]

\[\frac{- x - 16}{x + 4} > 0\]

\[\frac{x + 16}{x + 4} < 0\]

\[(x + 16)(x + 4) < 0\]

\[x \in ( - 16;\ - 4).\]

\[\textbf{б)}\ \frac{3 - x}{x - 2} < 1\]

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

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

\[\frac{- 2x + 5}{x - 2} < 0\]

\[\frac{2x - 5}{x - 2} > 0\]

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

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

\[\textbf{в)}\ \frac{7x - 1}{x} > 5\]

\[\frac{7x - 1}{x} - 5^{\backslash x} > 0\]

\[\frac{7x - 1 - 5x}{x} > 0\]

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

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

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

\[\textbf{г)}\ \frac{6 - 2x}{x + 4} > 3\]

\[\frac{6 - 2x}{x + 4} - 3^{\backslash x + 4} > 0\]

\[\frac{6 - 2x - 3x - 12}{x + 4} > 0\]

\[\frac{- 5x - 6}{x + 4} > 0\]

\[\frac{5x + 6}{x + 4} < 0\]

\[5 \cdot (x + 1,2)(x + 4) < 0\]

\[x \in ( - 4; - 1,2).\]

\[\boxed{\text{337.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\Longrightarrow не\ подходят.\]

\[Ответ:a = \frac{1}{2};\ \ a = 2.\]