Question

show me how to solve for x when 1/(x+1)^2 = [x/(x+1) - 2]/(x-1)

Answer 1

bunch of algebra, i will try to write it

Answer 2

ik thanks

Answer 3

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

Answer 4

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

Answer 5

wait how did you start out like that

Answer 6

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

Answer 7

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

Answer 8

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