Решебник по алгебре 9 класс Макарычев Задание 295

 

\[\boxed{\text{295}\text{\ (295)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ {x^{2}}^{\backslash x} + x^{\backslash x} - 9^{\backslash x} = \frac{9}{x}\]

\[x^{3} + x^{2} - 9x - 9 = 0;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[x \neq 0\]

\[x^{2}(x + 1) - 9 \cdot (x + 1) = 0\]

\[\left( x^{2} - 9 \right)(x + 1) = 0\]

\[(x - 3)(x + 3)(x + 1) = 0\]

\[x_{1} = 3,\ \ x_{2} = - 3,\ \ \]

\[x_{3} = - 1;\]

\[y(3) = \frac{9}{3} = 3 \Longrightarrow (3;3);\]

\[y( - 3) = \frac{9}{- 3} = - 3 \Longrightarrow ( - 3;\ - 3);\]

\[y( - 1) = \frac{9}{- 1} = - 9 \Longrightarrow ( - 1;\ - 9).\]

\[\textbf{б)}\ x^{2} + 6x - 4 = \frac{24}{x}\ \ \ \ \ | \cdot x;\ \]

\[\ x \neq 0\]

\[x^{3} + 6x^{2} - 4x - 24 = 0\]

\[x^{2}(x + 6) - 4 \cdot (x + 6) = 0\]

\[\left( x^{2} - 4 \right)(x + 6) = 0\]

\[(x - 2)(x + 2)(x + 6) = 0\]

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

\[x_{3} = - 6;\]

\[y(2) = \frac{24}{2} = 12 \Longrightarrow (2;12);\]

\[y( - 2) = \frac{24}{- 2} = - 12 \Longrightarrow\]

\[\Longrightarrow ( - 2;\ - 12)\mathbf{;}\]

\[y( - 6) = \frac{24}{- 6} = - 4 \Longrightarrow ( - 6; - 4).\]

\[\boxed{\text{295.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{x - 21}{x + 7} < 0\]

\[(x - 21)(x + 7) < 0\]

\[x \in ( - 7;21);\]

\[\textbf{б)}\ \frac{x + 4,7}{x - 7,2} > 0\]

\[(x + 4,7)(x - 7,2) > 0\]

\[x \in ( - \infty;\ - 4,7) \cup (7,2; + \infty).\]

\[\textbf{в)}\ \frac{6x + 1}{3 + x} > 0\]

\[(6x + 1)(x + 3) > 0\]

\[6 \cdot (x + 3)\left( x + \frac{1}{6} \right) > 0\]

\[x \in ( - \infty;\ - 3) \cup \left( - \frac{1}{6}; + \infty \right).\]

\[\textbf{г)}\ \frac{5x}{4x - 12} < 0\]

\[5x(4x - 12) < 0\]

\[20x(x - 3) < 0\]

\[x \in (0;3).\]