Đáp án:
$\begin{array}{l}
y = \frac{{{x^2} + x + 1}}{{{x^2} - 1}}\\
\Rightarrow {\rm{y}}{{\rm{x}}^2} - y = {x^2} + x + 1\\
\Rightarrow \left( {y - 1} \right){x^2} - x - y - 1 = 0\\
Để\,pt\,có\,nghiệm\, \Rightarrow \Delta \ge 0\\
\Rightarrow 1 - 4\left( {y - 1} \right).\left( { - y - 1} \right) \ge 0\\
\Rightarrow 1 + 4\left( {y - 1} \right)\left( {y + 1} \right) \ge 0\\
\Rightarrow 1 + 4{y^2} - 4 \ge 0\\
\Rightarrow 4{y^2} \ge 3\\
\Rightarrow {y^2} \ge \frac{3}{4}\\
\Rightarrow \left[ \begin{array}{l}
y \ge \frac{{\sqrt 3 }}{2}\\
y \le \frac{{ - \sqrt 3 }}{2}
\end{array} \right.
\end{array}$
