Question

find all real x such that \[\frac{1}{x} + \frac{1}{1-x} \gt 0\]find all real x such that \[\frac{1}{x} + \frac{1}{1-x} \gt 0\]@Mathematics

Answer 1

I would, but please don't post in other folks questions asking such things! :P

Answer 2

\[\frac{1}{x}+\frac{1}{1-x}=\frac{1-x+x}{x(1-x)}=\frac{1}{x(1-x)}\]for this to be >0, the denominator must be >0. therefore:\[x(1-x)>0\]therefore either both x and (1-x) have to be positive (but not zero), or both have to be negative (but not zero). the first restriction gives us:\[x > 0, x < 1\]and the second gives:\[x<0,x>1\]since the second restriction gives us a non-viable solution, the answer must be:\[0 < x< 1\]

Answer 3

YAY THANKS FOR THAT :) oops caps sorry :P