Решебник по алгебре 9 класс Макарычев Задание 187

 

\[\boxed{\text{187\ (187).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = \frac{8x - 7}{x} = \frac{8x}{x} - \frac{7}{x} = 8 - \frac{7}{x}\]

\[\frac{7}{x}\text{\ \ }должно\ быть\ целым \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} \frac{7}{x} = 7\ \ \\ \frac{7}{x} = 1\ \ \ \\ \frac{7}{x} = - 1 \\ \frac{7}{x} = - 7 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = 1\ \ \\ x = 7\ \ \ \\ x = - 7 \\ x = - 1 \\ \end{matrix} \right.\ \]

\[y(1) = 8 - \frac{7}{1} = 1;\]

\[y(7) = 8 - \frac{7}{7} = 8 - 1 = 7;\]

\[y( - 7) = 8 + \frac{7}{7} = 8 + 1 = 9;\]

\[y( - 1) = 8 + \frac{7}{1} = 8 + 7 = 15;\]

\[Ответ:(1;1),\ (7;7),\ ( - 7;9),\ \]

\[( - 1;15).\]

\[\boxed{\text{187.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ f(x);\ \ g(x) - четные\ \]

\[функции.\]

\[f( - x) = f(x);\ \ g( - x) = g(x)\]

\[y( - x) = f( - x) \cdot g( - x) =\]

\[= f(x) \cdot g(x) = y(x).\]

\[\textbf{б)}\ f(x);g(x) - нечетные\ \]

\[функции.\]

\[f( - x) = - f(x);g( - x) = - g(x)\]

\[y( - x) = - f(x) \cdot \left( - g(x) \right) =\]

\[= f(x) \cdot g(x) = y(x).\]

\[\textbf{в)}\ f(x) - четная\ функция;\]

\[g(x) - нечетная\ функция.\]

\[f( - x) = f(x);\ \ g( - x) = - g(x)\]

\[y( - x) = f( - x) \cdot \left( g( - x) \right) =\]

\[= f(x) \cdot \left( - g(x) \right) = - y(x).\]