Question

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

Answer 1

$\frac{x^2+x+1}{|x+1|}>0$ (i)
$\therefore$ $|x+1|>0,\forall xϵR-\{-1\}$
$\therefore$ (i) becomes $x^2+x+1>0$ $x\ne 1$
$\because$ $x^2+x+1>0,\forall xϵR$ (as its $a>0$ and $D<0$)
$\therefore$ Solution of (i) is $xϵR-\{-1\}$