ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 1124

\[\boxed{\text{1124.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

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

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

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

\[Значит:\ \ g(2) = g( - 2).\]

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

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

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

\[= \frac{- 2}{4 + 5} = - \frac{2}{9}.\]

\[Значит:\ \ g(2) > g( - 2).\]

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

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

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

\[= \frac{2}{4 + 5} = \frac{2}{9}.\]

\[Значит:\ \ g(2) < g( - 2).\]