Question

Solve the following inequality:
$\frac{x-1}{x+1}<x$

Answer 1

$\frac{x-1}{x+1}<x$

$\frac{x-1}{x+1}-x<0$

$\frac{x-1-x^2-x}{x+1}<0$

$\frac{x^2+1}{x+1}>0$

$\frac{1}{x+1}>0$

$xϵ(-1,\infty )$