ГДЗ по алгебре 9 класс Макарычев Задание 53

\[\boxed{\text{53\ (53).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ g(x) = \frac{1}{x^{2} + 5}\ \]

\[g(2) = \frac{1}{2^{2} + 5} = \frac{1}{4 + 5} = \frac{1}{9};\]

\[g( - 2) = \frac{1}{( - {2)}^{2} + 5} = \frac{1}{9}.\]

\[Значит:\ \ \]

\[g(2) = g( - 2).\]

\[\textbf{б)}\ g(x) = \frac{x}{x^{2} + 5}\ \]

\[g(2) = \frac{2}{2^{2} + 5} = \frac{2}{4 + 5} = \frac{2}{9};\]

\[g( - 2) = \frac{- 2}{( - {2)}^{2} + 5} = \frac{- 2}{4 + 5} =\]

\[= - \frac{2}{9}.\]

\[Значит:\ \ \]

\[g(2) > g( - 2).\]

\[\textbf{в)}\ g(x) = \frac{- x}{x^{2} + 5}\ \]

\[g(2) = \frac{- 2}{2^{2} + 5} = \frac{- 2}{4 + 5} = - \frac{2}{9};\]

\[g( - 2) = \frac{2}{( - {2)}^{2} + 5} = \frac{2}{4 + 5} = \frac{2}{9}.\]

\[Значит:\ \]

\[g(2) < g( - 2).\]