Question

pe 4 va rog frumos ....urgen pentru maine dau coroana

Answer 1

$\displaystyle E= \frac{1}{x-1}+ \frac{2x+3}{x^2+2x+1}- \frac{1}{(x+1)^2}- \frac{3x-2}{x^2-1}= \\ \\~~~ \\ = \frac{2x+3}{x^2+2x+1}- \frac{1}{(x+1)^2}+ \frac{1}{x-1}- \frac{3x-2}{x^2-1}= \\ \\ \\ = \frac{2x+3}{(x+1)^2}- \frac{1}{(x+1)^2}+ \frac{x+1}{(x+1)(x-1)}- \frac{3x-2}{x^2-1}= \\ \\ \\ = \frac{2x+2}{(x+1)^2}+ \frac{x+1}{x^2-1}- \frac{3x-2}{x^2-1}= \\ \\ \\ = \frac{2(x+1)}{(x+1)^2}+ \frac{x+1-3x+2}{x^2-1} =$

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