Решебник по алгебре 9 класс Макарычев Задание 237

 

\[\boxed{\text{237\ (237).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ y = x|x|\]

\[При\ x < 0:\]

\[y = - x \cdot x = - x^{2};\]

\[При\ x \geq 0:\]

\[y = x^{2}.\]

\[\textbf{б)}\ y = - \frac{x^{3}}{|x|};\ \ \ \ x \neq 0\]

\[x = 0 - выколотая\ точка.\]

\[При\ x < 0:\]

\[y = \frac{x^{3}}{x} = x^{2};\]

\[При\ x > 0:\]

\[y = - \frac{x^{3}}{x} = - x^{2}\text{.\ }\]

\[\boxed{\text{237.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{a + 1}{a - 2} + \frac{a - 4}{a + 1} = \frac{3a + 3}{a^{2} - a - 2}\]

\[2a^{2} - 7a + 6 = 0\]

\[ОДЗ:\ \ \ a \neq 2;\ - 1.\]

\[D = 7^{2} - 4 \cdot 2 \cdot 6 = 49 - 48 = 1\]

\[a = \frac{7 \pm 1}{4},\ \ \]

\[a_{1} = 1,5;\ \ a_{2} = 2.\]

\[Ответ:при\ a = 1,5.\]

\[\textbf{б)}\ \frac{3a - 5}{a^{2} - 1} - \frac{6a - 5}{a - a^{2}} = \frac{3a + 2}{a^{2} + a}\]

\[\frac{3a - 5}{(a - 1)(a + 1)} - \frac{6a - 5}{a(1 - a)} =\]

\[= \frac{3a + 2}{a(a + 1)}\]

\[\frac{6a^{2} - 3a - 3}{a(a - 1)(a + 1)} = 0\]

\[6a^{2} - 3a - 3 = 0\]

\[ОДЗ:\ \ \ a \neq 0;\ - 1;1.\]

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

\[D = 1 + 4 \cdot 2 = 9\]

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

\[Учитывая\ ОДЗ:a = - \frac{1}{2}.\]

\[Ответ:a = - 0,5.\]