Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 17

\[1)\ y(x) = f_{1}(x) \bullet f_{2}(x);\ \ \ \]

\[g(x) = \frac{f_{1}(x)}{f_{2}(x)}:\]

\[y( - x) = f_{1}( - x) \bullet f_{2}( - x) =\]

\[= - f_{1}(x) \bullet \left( - f_{2}(x) \right) =\]

\[= f_{1}(x) \bullet f_{2}(x) = y(x);\]

\[g( - x) = \frac{f_{1}( - x)}{f_{2}( - x)} = \frac{- f_{1}(x)}{- f_{2}(x)} =\]

\[= \frac{f_{1}(x)}{f_{2}(x)} = g(x).\]

\[Что\ и\ требовалось\ доказать.\]

\[2)\ y(x) = f_{1}(x) \bullet f_{2}(x);\ \ \ \]

\[g(x) = \frac{f_{1}(x)}{f_{2}(x)}:\]

\[y( - x) = f_{1}( - x) \bullet f_{2}( - x) =\]

\[= - f_{1}(x) \bullet f_{2}(x) = - y(x);\]

\[g( - x) = \frac{f_{1}( - x)}{f_{2}( - x)} = \frac{- f_{1}(x)}{f_{2}(x)} = - g(x).\]

\[Что\ и\ требовалось\ доказать.\]