Решебник по алгебре 9 класс Макарычев Задание 53

 

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

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

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

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

\[Значит:\ \ \]

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

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

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

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

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

\[Значит:\ \ \]

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

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

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

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

\[Значит:\ \]

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

\[\boxed{\text{53.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x^{2} - 8x + q = 0;\ \ \]

\[\ x_{1} - x_{2} = 16\]

\[по\ теореме\ Виета:\]

\(\left\{ \begin{matrix} x_{1} + x_{2} = 8 \\ x_{1} \cdot x_{2} = q \\ \end{matrix} \right.\ \text{\ \ \ \ }\)

\[из\ условия:\ \]

\[x_{1} - x_{2} = 16;\]

\[x_{1} = 16 + x_{2};\ \ \]

\[подставим\]

\[Ответ:q = - 48.\]

\[\textbf{б)}\ x^{2} - 7x + q = 0;\ \ \]

\[\ x_{1}^{2} + x_{2}^{2} = 29\]

\[x_{1}^{2} + 2x_{1}x_{2} + x_{2}^{2} - 2x_{1}x_{2} = 29\]

\[\left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2} = 29\]

\[по\ теореме\ Виета:\ \]

\[\left\{ \begin{matrix} x_{1} + x_{2} = 7 \\ x_{1} \cdot x_{2} = q \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[подставим:\]

\[7^{2} - 2 \cdot q = 29\]

\[2q = 49 - 29 = 20\ \ \ |\ \ :2\]

\[q = 10\]

\[Ответ:q = 10.\]