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