Решебник по алгебре 9 класс Макарычев Задание 459

 

\[\boxed{\text{459\ (459).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Пусть\ x - длина\ участка,\ \]

\[а\ y - его\ ширина.\ \]

\[Площадь\ прямоугольника:\]

\[S = xy;\ \ по\ условию\ \]

\[она\ = 2400\ м^{2}\text{.\ \ }\]

\[Периметр\ участка\ 2 \cdot (x + y) =\]

\[= 200.\]

\[Составим\ систему\ уравнений:\]

\[\left\{ \begin{matrix} xy = 2400\ \ \ \ \ \ \ \ \ \ \ \ \\ 2 \cdot (x + y) = 200 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} xy = 2400\ \ \\ x + y = 100 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 100 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ y(100 - y) = 2400 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 100 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 100y + 2400 = 0 \\ \end{matrix} \right.\ \]

\[y^{2} - 100y + 2400 = 0\]

\[D_{1} = 2500 - 2400 = 100\]

\[y_{1} = 50 + 10 = 60;\ \]

\[\ y_{2} = 50 - 10 = 40.\]

\[\left\{ \begin{matrix} y_{1} = 60 \\ x_{1} = 40 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} y_{2} = 40 \\ x_{2} = 60 \\ \end{matrix}. \right.\ \]

\[Ответ:длина\ 60\ м,\ \]

\[ширина\ 40\ м.\]

\(\boxed{\text{459.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)

\[\frac{x - 1}{x + 2} - \frac{1 - x}{x^{2} + 3x + 2} =\]

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

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

\[= \frac{x^{2} - 1 - 1 + x}{(x + 1)(x + 2)} =\]

\[= \frac{x^{2} + x - 2}{(x + 1)(x + 2)};\]

\[x^{2} + 3x + 2 = (x + 2)(x + 1)\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = 2\]

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

\[x^{2} + x - 2 = (x + 2)(x - 1)\]

\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 2\]

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

\[\Longrightarrow \frac{(x + 2)(x - 1)}{(x + 1)(x + 2)} = \frac{x - 1}{x + 1}.\]