Đề: $-3 < \frac{x^2-3x-1}{x^2+x+1}<3$
$↔$ \(\left[ \begin{array}{l}\frac{x^2-3x-1}{x^2+x+1}>-3\\\frac{x^2-3x-1}{x^2+x+1}<3\end{array} \right.\)
$↔ $\(\left[ \begin{array}{l}\frac{x^2-3x-1+3(x^2+x+1)}{x^2+x+1}>0\\\frac{x^2-3x-1-3(x^2+x+1)}{x^2+x+1}<0\end{array} \right.\)
$↔$ \(\left[ \begin{array}{l}4x^2+2>0 (ld)\\-2x^2-6x-4<0\end{array} \right.\)
$↔ x^2 +3x +2>0$
$↔ (x+1)(x+2)>0$
$↔$ \(\left[ \begin{array}{l}x>-1\\x<-2\end{array} \right.\)
