## anonymous 4 years ago show me how to solve for x when 1/(x+1)^2 = [x/(x+1) - 2]/(x-1)

1. anonymous

bunch of algebra, i will try to write it

2. anonymous

ik thanks

3. anonymous

$\frac{x}{x+1}-2=\frac{-x-2}{x+1}$ is a start

4. anonymous

so right hand sides is $\frac{-x-2}{(x+1)(x-1)}$

5. anonymous

wait how did you start out like that

6. anonymous

$\frac{1}{(x+1)^2}=\frac{-x-2}{(x+1)(x-1)}$ $\frac{1}{x+1}=\frac{-x-2}{x-1}$ $x-1=(x+1)(-x-2)$ $x-1=-x^2-3x-2$ $x^2+3x+1=0$

7. anonymous

$\frac{x}{x+1}-2=\frac{x}{x+1}-\frac{2(x+1)}{x+1}$ etc

8. anonymous

still don't get thte first step talk about moronic what the flutter