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Trả lời:
$f(x)=\dfrac{2}{x^2-5x+9}=\dfrac{2}{x^2-2.x.\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{11}{4}}=\dfrac{2}{\bigg{(}x-\dfrac{5}{2}\bigg{)}^2+\dfrac{11}{4}}$
Ta có: $\bigg{(}x-\dfrac{5}{2}\bigg{)}^2+\dfrac{11}{4}\geq \dfrac{11}{4}$
$⇒f(x)=\dfrac{2}{\bigg{(}x-\dfrac{5}{2}\bigg{)}^2+\dfrac{11}{4}}\leq \dfrac{2}{\dfrac{11}{4}}=\dfrac{8}{11}$
Dấu "=" xảy ra khi: $x-\dfrac{5}{2}=0⇔x=\dfrac{5}{2}$
Vậy $max_{f(x)}=\dfrac{8}{11}⇔x=\dfrac{5}{2}$.
