Предмет: Алгебра, автор: ashihminko

Помогите, срочно!!! Написать решение и ответ.даю 20 балов

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Ответы

Автор ответа: ajax1332

Ответ:

$\frac{x - y}{x} - \frac{5y}{x^{2}} \cdot \frac{x^{2} - xy}{5y} = \frac{x - y}{x} - \frac{5y}{x^{2}} \cdot \frac{x(x - y)}{5y} = \frac{x - y}{x} - \frac{x - y}{x} = 0\\\\(\frac{4a}{2 - a} - a):\frac{a + 2}{a - 2} = (\frac{4a - 2a + a^{2}}{2 - a}):\frac{a + 2}{a - 2} = (\frac{a(2 + a)}{2 - a}):\frac{a + 2}{a - 2} = -\frac{a(a + 2)}{a - 2} \cdot \frac{a - 2}{a + 2} = -a\\\\$ $\frac{a^{2} - 9}{2a^{2} + 1} \cdot (\frac{(6a + 1)(a + 3) + (6a - 1)(a - 3)}{a^{2} - 9}) = \frac{a^{2} - 9}{2a^{2} + 1} \cdot (\frac{(6a^{2} + 18a + a + 3 + 6a^{2} - 18a - a + 3}{a^{2} - 9}) = \frac{a^{2} - 9}{2a^{2} + 1} \cdot (\frac{12a^{2} + 6}{a^{2} - 9}) = \frac{a^{2} - 9}{2a^{2} + 1} \cdot \frac{6(2a^{2} + 1)}{a^{2} - 9} = 6\\$ $\frac{a - 2}{4a^{2} + 16a + 16} : (\frac{a}{2a - 4} - \frac{a^{2} + 4}{2a^{2} - 8} - \frac{2}{a^{2} + 2a}) = \frac{a - 2}{4(a + 2)^{2}} : (\frac{a}{2(a - 2)} - \frac{a^{2} + 4}{2(a - 2)(a + 2)} - \frac{2}{a(a + 2)}) = \frac{a - 2}{4(a + 2)^{2}} : (\frac{a*a(a + 2) - a(a^{2} + 4) - 4(a - 2)}{2a(a - 2)(a + 2)}}) = \frac{a - 2}{4(a + 2)^{2}} : (\frac{a^{3} + 2a^{2} - a^{3} - 4a - 4a + 8)}{2a(a - 2)(a + 2)}}) = \frac{a - 2}{4(a + 2)^{2}} : (\frac{2(a - 2)^{2}}{2a(a - 2)(a + 2)}})$ $= \frac{a - 2}{4(a + 2)^{2}} : (\frac{(a - 2)}{a(a + 2)}}) = \frac{a - 2}{4(a + 2)^{2}} \cdot (\frac{a(a + 2)}{(a - 2)}}}) = \frac{a}{4(a + 2)}}}$ $\\(\frac{x - 2y}{x^{2} + 2xy} - \frac{1}{x^{2} - 4y^{2}}} : \frac{x + 2y}{(2y - x)^{2}})\cdot\frac{(x + 2y)^{2}}{4y^{2}} = (\frac{x - 2y}{x(x + 2y} - \frac{1}{(x - 2y)(x + 2y)}} : -\frac{x + 2y}{(x - 2y)(x - 2y)})\cdot\frac{(x + 2y)^{2}}{4y^{2}} = (\frac{x - 2y}{x(x + 2y} - \frac{1}{(x - 2y)(x + 2y)}} \cdot -\frac{(x - 2y)(x - 2y)}{x + 2y})\cdot\frac{(x + 2y)^{2}}{4y^{2}} = (\frac{x - 2y}{x(x + 2y)} - \frac{-x + 2y}{(x + 2y)}})\cdot\frac{(x + 2y)^{2}}{4y^{2}} =$