Đáp án:
\[f{\left( x \right)_{\max }} = \frac{8}{{11}} \Leftrightarrow x = \frac{5}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = \frac{2}{{{x^2} - 5x + 9}}\\
{x^2} - 5x + 9 = \left( {{x^2} - 2.x.\frac{5}{2} + \frac{{25}}{4}} \right) + \frac{{11}}{4} = {\left( {x - \frac{5}{2}} \right)^2} + \frac{{11}}{4} \ge \frac{{11}}{4},\,\,\,\forall x\\
\Rightarrow f\left( x \right) = \frac{2}{{{x^2} - 5x + 9}} \le \frac{2}{{\frac{{11}}{4}}} = \frac{8}{{11}}\\
\Rightarrow f{\left( x \right)_{\max }} = \frac{8}{{11}} \Leftrightarrow x = \frac{5}{2}
\end{array}\)
