Đáp án:
\[\left\{ \begin{array}{l}
{P_{\min }} = - 1 \Leftrightarrow x = 2\\
{P_{\max }} = 9 \Leftrightarrow x = - \frac{1}{2}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = \frac{{{x^2} - 8x + 7}}{{{x^2} + 1}}\\
P + 1 = \frac{{{x^2} - 8x + 7}}{{{x^2} + 1}} + 1 = \frac{{{x^2} - 8x + 7 + {x^2} + 1}}{{{x^2} + 1}}\\
= \frac{{2{x^2} - 8x + 8}}{{{x^2} + 1}} = \frac{{2\left( {{x^2} - 4x + 4} \right)}}{{{x^2} + 1}} = \frac{{2{{\left( {x - 2} \right)}^2}}}{{{x^2} + 1}}\\
\frac{{2{{\left( {x - 2} \right)}^2}}}{{{x^2} + 1}} \ge 0,\,\,\,\,\forall x \Rightarrow P + 1 \ge 0,\,\,\,\,\forall x \Leftrightarrow P \ge - 1,\,\,\,\,\forall x\\
\Rightarrow {P_{\min }} = - 1 \Leftrightarrow x = 2\\
P - 9 = \frac{{{x^2} - 8x + 7}}{{{x^2} + 1}} - 9 = \frac{{\left( {{x^2} - 8x + 7} \right) - 9\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}\\
= \frac{{ - 8{x^2} - 8x - 2}}{{{x^2} + 1}} = \frac{{ - 2\left( {4{x^2} + 4x + 1} \right)}}{{{x^2} + 1}} = \frac{{ - 2{{\left( {2x + 1} \right)}^2}}}{{{x^2} + 1}}\\
\frac{{{{\left( {2x + 1} \right)}^2}}}{{{x^2} + 1}} \ge 0,\,\,\,\,\,\,\forall x \Rightarrow \frac{{ - 2{{\left( {2x + 1} \right)}^2}}}{{{x^2} + 1}} \le 0,\,\,\,\,\forall x \Rightarrow P - 9 \le 0,\,\,\,\,\forall x\\
\Rightarrow P \le 9,\,\,\,\,\forall x\\
\Rightarrow {P_{\max }} = 9 \Leftrightarrow 2x + 1 = 0 \Leftrightarrow x = - \frac{1}{2}
\end{array}\)
Vậy \(\left\{ \begin{array}{l}
{P_{\min }} = - 1 \Leftrightarrow x = 2\\
{P_{\max }} = 9 \Leftrightarrow x = - \frac{1}{2}
\end{array} \right.\)
