Предмет: Алгебра, автор: Lqdqiaq

Помогите пожалуйста срочно

Приложения:

Ответы

Автор ответа: Pelmeshka83

Ответ:

$3)( \frac{a}{a + b} + \frac{b}{a - b} + \frac{2ab}{ {b}^{2} - {a}^{2} } ) \times \frac{a}{a - b} + ( \frac{b}{b - a} + \frac{2ab}{ {a}^{2} - {b}^{2} } ) = \frac{ - a(b - a) + b(a + b) - 2ab}{ {b}^{2} - {a}^{2} } \times \frac{a}{a - b} + \frac{ - b( {a}^{2} - {b}^{2}) + 2ab }{ {a}^{2} - {b}^{2} } = \frac{ {a}^{2} - ab + ab + {b}^{2} - 2ab }{(b - a)(b + a)} \times \frac{a}{a - b} + \frac{ {b}^{3} - {a}^{2}b + 2ab }{(a - b)(a + b)} = \frac{ {(a - b)}^{2} }{ - (a - b)(a + b)} \times \frac{a}{a - b} + \frac{b( {b}^{2} + 2a - {a}^{2}) }{(a - b)(a + b)} = - \frac{a(a - b) + b( {b}^{2} + 2a - {a}^{2} ) }{(a + b)(a - b)} = \frac{ab - {a}^{2} + {b}^{3} + 2ab - {a}^{2} b }{(a + b)(a - b)} = \frac{ {b}^{3} - {a}^{2}b + 3ab - {a}^{2} }{(a + b)(a - b)}$

$4)( \frac{x}{ {x}^{2} - {y}^{2} } - \frac{x}{ {(x - y)}^{2} } ) \times \frac{ {y {}^{2} - 2xy + {x}^{2} }^{} }{2x} + \frac{y}{x + y} = \frac{x(x - y) - x(x + y)}{(x + y) {(x - y)}^{2} } \times \frac{ {(y - x)}^{2} }{2x} + \frac{y}{x + y} = \frac{ {x}^{2} - xy - {x}^{2} - xy }{(x + y) {(x - y)}^{2} } \times \frac{ {(y - x)}^{2} }{2x} + \frac{y}{x + y} = \frac{ - 2xy \times (x - y) {}^{2} }{(x + y) {(x - y)}^{2} \times 2x } + \frac{ y}{x + y} = \frac{ - y}{x + y} + \frac{y}{x + y} = 0$