Đáp án:
$\begin{array}{l}
y = \dfrac{{x + 1}}{{{x^2} - x + 1}}\\
\Rightarrow y.{x^2} - y.x + y = x + 1\\
\Rightarrow y.{x^2} - \left( {y + 1} \right).x + y - 1 = 0\\
Dk\,có\,nghiệm\\
\Rightarrow \Delta \ge 0\\
\Rightarrow {\left( {y + 1} \right)^2} - 4.y.\left( {y - 1} \right) \ge 0\\
\Rightarrow {y^2} + 2y + 1 - 4{y^2} + 4y \ge 0\\
\Rightarrow 3{y^2} - 6y - 1 \le 0\\
\Rightarrow {y^2} - 2y - \dfrac{1}{3} \le 0\\
\Rightarrow {\left( {y - 1} \right)^2} \le 1 + \dfrac{1}{3} = \dfrac{4}{3}\\
\Rightarrow 1 - \dfrac{{2\sqrt 3 }}{3} \le y \le 1 + \dfrac{{2\sqrt 3 }}{3}\\
\Rightarrow \left\{ \begin{array}{l}
\min y = 1 - \dfrac{{2\sqrt 3 }}{3}\\
\max y = 1 + \dfrac{{2\sqrt 3 }}{3}
\end{array} \right.
\end{array}$
