Đáp án:
$\begin{array}{l}
P = \dfrac{{{x^2} + 1}}{{{x^2} - x + 1}}\\
\Rightarrow P.{x^2} - P.x + P = {x^2} + 1\\
\Rightarrow \left( {P - 1} \right).{x^2} - P.x + P - 1 = 0\\
\Rightarrow \Delta \ge 0\\
\Rightarrow {P^2} - 4.\left( {P - 1} \right).\left( {P - 1} \right) \ge 0\\
\Rightarrow {P^2} - 4{\left( {P - 1} \right)^2} \ge 0\\
\Rightarrow \left( {P - 2P + 2} \right)\left( {P + 2P - 2} \right) \ge 0\\
\Rightarrow \left( {2 - P} \right)\left( {3P - 2} \right) \ge 0\\
\Rightarrow \left( {P - 2} \right)\left( {3P - 2} \right) \le 0\\
\Rightarrow \dfrac{2}{3} \le P \le 2\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:P = \dfrac{2}{3}\,khi:x = - 1\\
GTLN:P = 2\,khi:x = 1
\end{array} \right.
\end{array}$
