Решите любые 2 номера,ПЖ!!!!

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12 просмотров

Решите любые 2 номера,ПЖ!!!!


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Алгебра (15 баллов) | 12 просмотров
Дан 1 ответ
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Правильный ответ

79.
1)
\frac{3x}{x^3-y^3}* \frac{x^2+xy+y^2}{x+y}= \frac{3x}{(x-y)(x^2+xy+y^2)}* \frac{x^2+xy+y^2}{x+y}= \frac{3x}{(x-y)(x+y)}

2)
\frac{3}{x-y}+ \frac{3x}{(x-y)(x+y)}= \frac{3(x+y)+3x}{(x-y)(x+y)}= \frac{3x+3y+3x}{(x-y)(x+y)}= \frac{6x+3y}{(x-y)(x+y)}= \frac{3(2x+y)}{(x-y)(x+y)}

3)
\frac{3(2x+y)}{(x-y)(x+y)} : \frac{2x+y}{x^2+2xy+y^2}= \frac{3(2x+y)}{(x-y)(x+y)}* \frac{(x+y)^2}{2x+y}= \frac{3(x+y)}{x-y}

4)
\frac{3(x+y)}{x-y}* \frac{3}{x+y}= \frac{9}{x-y}

85.
1)
\frac{x^2}{x+y}- \frac{x^3}{x^2+2xy+y^2}= \frac{x^2}{x+y}- \frac{x^3}{(x+y)^2}= \frac{x^2(x+y)-x^3}{(x+y)^2}= \frac{x^3+x^2y-x^3}{(x+y)^2}= \\ = \frac{x^2y}{(x+y)^2}

2)
\frac{x^2}{x^2-y^2}+ \frac{x}{y-x}= \frac{x^2}{(x-y)(x+y)}- \frac{x}{x-y}= \frac{x^2-x(x+y)}{(x-y)(x+y)}= \frac{x^2-x^2-xy}{(x-y)(x+y)}= \\ = -\frac{xy}{(x-y)(x+y)}

3)
\frac{x^2y}{(x+y)^2}:(- \frac{xy}{(x-y)(x+y)} )= \frac{x^2y}{(x+y)^2}*(- \frac{(x-y)(x+y)}{xy} )=- \frac{x(x-y)}{x+y}= \\ = \frac{x(y-x)}{x+y}

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

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