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Đáp án:

$MaxP = 2 \Leftrightarrow x = 1;MinP = \dfrac{2}{3} \Leftrightarrow x =  – 1$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
 + )P – \dfrac{2}{3} = \dfrac{{{x^2} + 1}}{{{x^2} – x + 1}} – \dfrac{2}{3}\\
 = \dfrac{{3\left( {{x^2} + 1} \right) – 2\left( {{x^2} – x + 1} \right)}}{{3\left( {{x^2} – x + 1} \right)}}\\
 = \dfrac{{{x^2} + 2x + 1}}{{3\left( {{x^2} – x + 1} \right)}}\\
 = \dfrac{{{{\left( {x + 1} \right)}^2}}}{{3\left( {{x^2} – x + 1} \right)}}
\end{array}$

Mà:

$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x + 1} \right)^2} \ge 0\\
{x^2} – x + 1 = {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0
\end{array} \right.,\forall x\\
 \Rightarrow \dfrac{{{{\left( {x + 1} \right)}^2}}}{{3\left( {{x^2} – x + 1} \right)}} \ge 0,\forall x\\
 \Rightarrow P – \dfrac{2}{3} \ge 0\\
 \Rightarrow P \ge \dfrac{2}{3}\\
 \Rightarrow MinP = \dfrac{2}{3} \Leftrightarrow {\left( {x + 1} \right)^2} = 0 \Leftrightarrow x =  – 1
\end{array}$

$\begin{array}{l}
 + )P – 2 = \dfrac{{{x^2} + 1}}{{{x^2} – x + 1}} – 2\\
 = \dfrac{{{x^2} + 1 – 2\left( {{x^2} – x + 1} \right)}}{{{x^2} – x + 1}}\\
 = \dfrac{{ – {x^2} + 2x – 1}}{{{x^2} – x + 1}}\\
 = \dfrac{{ – {{\left( {x – 1} \right)}^2}}}{{{x^2} – x + 1}}
\end{array}$

Mà:

$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x – 1} \right)^2} \ge 0\\
{x^2} – x + 1 = {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0
\end{array} \right.,\forall x\\
 \Rightarrow \dfrac{{ – {{\left( {x – 1} \right)}^2}}}{{3\left( {{x^2} – x + 1} \right)}} \le 0,\forall x\\
 \Rightarrow P – 2 \le 0\\
 \Rightarrow P \le 2\\
 \Rightarrow MaxP = 2 \Leftrightarrow {\left( {x – 1} \right)^2} = 0 \Leftrightarrow x = 1
\end{array}$

Vậy $MaxP = 2 \Leftrightarrow x = 1;MinP = \dfrac{2}{3} \Leftrightarrow x =  – 1$