Question

Solve
$|\frac{x^2-1}{x^2+x+1}|<1$.

Answer 1

Given,

$\left|\frac{x^2-1}{x^2+x+1}\right|<1$

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

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

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

$(x<-\frac{1}{2}orx>0)$ and $x>-2$

$-2<x<-\frac{1}{2}orx>0$