Предмет: Алгебра, автор: galianakipova9

Помогите решить пожалуйста. 581 2) и 582 2)

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Автор ответа: inblu

$(\frac{1}{x+1}- \frac{3}{x^3+1}+ \frac{3}{x^2-x+1})(x- \frac{2x-1}{x+1})= \\ =(\frac{1}{x+1}- \frac{3}{(x+1)(x^2-x+1)}+ \frac{3}{x^2-x+1})*\frac{x^2+x-2x+1}{x+1}= \\ =\frac{x^2-x+1-3+3x+3}{(x+1)(x^2-x+1)}*\frac{x^2-x+1}{x+1}=\frac{x^2+2x+1}{(x+1)(x^2-x+1)}*\frac{x^2-x+1}{x+1}= \frac{(x+1)^2}{(x+1)(x+1)}=1$

$(\frac{a-1}{3a+(a-1)^2}- \frac{1-2a+a^2}{a^3-1}- \frac{1}{a-1}): \frac{a^2+1}{1-a}= \\ =(\frac{a-1}{3a+a^2-2a+1}- \frac{1-2a+a^2}{(a-1)(a^2+a+1)}- \frac{1}{a-1}): \frac{a^2+1}{1-a}= \\ =(\frac{a-1}{a^2+a+1}- \frac{1-2a+a^2}{(a-1)(a^2+a+1)}- \frac{1}{a-1}): \frac{a^2+1}{1-a}= \\ =\frac{(a-1)^2-(1-2a+a^2)-(a^2+a+1)}{(a-1)(a^2+a+1)}: \frac{a^2+1}{1-a}= \\ =\frac{a^2-2a+1-1+2a-a^2-a^2-a-1)}{(a-1)(a^2+a+1)}* \frac{1-a}{a^2+1}=\frac{-a^2-a-1}{(a-1)(a^2+a+1)}* \frac{1-a}{a^2+1}=$
$=\frac{-(a^2+a+1)}{(a-1)(a^2+a+1)}* \frac{-(a-1)}{a^2+1}= \frac{1}{a^2+1}$

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