Страница 46 номер 2.3.35, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
При \(x=-2{,}5\), \(y=0{,}5\) найти значение выражения: \[\left(\frac{x-2y}{x^3+y^3}+\frac{y}{x^3-x^2y+xy^2}\right)\cdot\frac{x^3-xy^2}{x^2+y^2}+\frac{2y^2}{x^3+x^2y+xy^2+y^3}.\]
Разложим на множители:
\[x^3 + y^3 = (x + y)(x^2 - xy + y^2);\]
\[x^3 - x^2 y + xy^2 = x(x^2 - xy + y^2);\]
\[x^3 - xy^2 = x(x^2 - y^2) = x(x - y)(x + y);\]
\[x^3 + x^2 y + xy^2 + y^3 = x^2(x + y) + y^2(x + y) = (x + y)(x^2 + y^2).\]
1) \[\begin{aligned} &\frac{x - 2y}{x^3 + y^3} + \frac{y}{x^3 - x^2 y + xy^2} = \frac{x - 2y}{(x + y)(x^2 - xy + y^2)} + \frac{y}{x(x^2 - xy + y^2)} = {} \\ &= \frac{x(x - 2y) + y(x + y)}{x(x + y)(x^2 - xy + y^2)} = \frac{x^2 - xy + y^2}{x(x + y)(x^2 - xy + y^2)} = \frac{1}{x(x + y)};\end{aligned}\]
2) \[\frac{1}{x(x + y)} \cdot \frac{x^3 - xy^2}{x^2 + y^2} = \frac{1}{x(x + y)} \cdot \frac{x(x - y)(x + y)}{x^2 + y^2} = \frac{x - y}{x^2 + y^2};\]
3) \[\begin{aligned} &\frac{x - y}{x^2 + y^2} + \frac{2y^2}{x^3 + x^2 y + xy^2 + y^3} = \frac{x - y}{x^2 + y^2} + \frac{2y^2}{(x + y)(x^2 + y^2)} = {} \\ &= \frac{(x - y)(x + y) + 2y^2}{(x + y)(x^2 + y^2)} = \frac{x^2 - y^2 + 2y^2}{(x + y)(x^2 + y^2)} = {} \\ &= \frac{x^2 + y^2}{(x + y)(x^2 + y^2)} = \frac{1}{x + y}.\end{aligned}\]
При \(x = -2{,}5\), \(y = 0{,}5\) получаем \[\frac{1}{x + y} = \frac{1}{-2{,}5 + 0{,}5} = \frac{1}{-2} = -0{,}5.\]
Ответ: \(-0{,}5\).
При \(x=-2{,}5\), \(y=0{,}5\) найти значение выражения: \[\left(\frac{x-2y}{x^3+y^3}+\frac{y}{x^3-x^2y+xy^2}\right)\cdot\frac{x^3-xy^2}{x^2+y^2}+\frac{2y^2}{x^3+x^2y+xy^2+y^3}.\]
Разложим на множители многочлены, входящие в данное выражение:
\[x^3 + y^3 = (x + y)(x^2 - xy + y^2);\]
\[x^3 - x^2 y + xy^2 = x(x^2 - xy + y^2);\]
\[x^3 - xy^2 = x(x^2 - y^2) = x(x - y)(x + y);\]
\[x^3 + x^2 y + xy^2 + y^3 = x^2(x + y) + y^2(x + y) = (x + y)(x^2 + y^2).\]
1) \[\begin{aligned} &\frac{x - 2y}{x^3 + y^3} + \frac{y}{x^3 - x^2 y + xy^2} = \frac{x - 2y}{(x + y)(x^2 - xy + y^2)} + \frac{y}{x(x^2 - xy + y^2)} = {} \\ &= \frac{x(x - 2y) + y(x + y)}{x(x + y)(x^2 - xy + y^2)} = \frac{x^2 - xy + y^2}{x(x + y)(x^2 - xy + y^2)} = \frac{1}{x(x + y)};\end{aligned}\]
2) \[\frac{1}{x(x + y)} \cdot \frac{x^3 - xy^2}{x^2 + y^2} = \frac{1}{x(x + y)} \cdot \frac{x(x - y)(x + y)}{x^2 + y^2} = \frac{x - y}{x^2 + y^2};\]
3) \[\begin{aligned} &\frac{x - y}{x^2 + y^2} + \frac{2y^2}{x^3 + x^2 y + xy^2 + y^3} = \frac{x - y}{x^2 + y^2} + \frac{2y^2}{(x + y)(x^2 + y^2)} = {} \\ &= \frac{(x - y)(x + y) + 2y^2}{(x + y)(x^2 + y^2)} = \frac{x^2 - y^2 + 2y^2}{(x + y)(x^2 + y^2)} = {} \\ &= \frac{x^2 + y^2}{(x + y)(x^2 + y^2)} = \frac{1}{x + y}.\end{aligned}\]
При \(x = -2{,}5\) и \(y = 0{,}5\) значение полученной дроби равно \[\frac{1}{x + y} = \frac{1}{-2{,}5 + 0{,}5} = \frac{1}{-2} = -0{,}5.\]
Ответ: \(-0{,}5\).
