Помогите Пожалуйста!!!!!! Срочно!!!!. Created by Bozhena127. algebra-ru

Помогите Пожалуйста!!!!!! Срочно!!!!...

маслёнок95 ((x(3x+9)-3(x-1))/(x²-х))*((х²-х)/(3х+3))=((3х²+9х-3х+3)/(х²-х))*((х²-х)/(3(х+1))= ((3х²+6х+3)/(х²-х))*((х²-х)/(3(х+1))= (3(х²+2х+1)/(х²-х))*((х²-х)/(3(х+1))= (3(х+1)²/(х²-х))*((х²-х)/(3(х+1))=х+1

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Sanyasha $( \frac{3x+9}{x^2-1}- \frac{3}{x^2+x} ): \frac{3x+3}{x^2-x}= ( \frac{3x+9}{(x-1)(x+1)}- \frac{3}{x(x+1)} ): \frac{3x+3}{x^2-x}= \\ ( \frac{3x^2+9x}{x(x-1)(x+1)}- \frac{3x-3}{x(x+1)(x-1)} ): \frac{3x+3}{x^2-x}= \frac{3x^2+9x-3x+3}{x(x-1)(x+1)} : \frac{3x+3}{x^2-x}= \\ \frac{3x^2+6x+3}{x(x-1)(x+1)} : \frac{3x+3}{x^2-x}= \frac{3(x-1)^2}{x(x-1)(x+1)} * \frac{x(x-1)}{3(x+1)}= \frac{(x-1)^2}{(x+1)^2}$

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