Đáp án:
$minP = \dfrac{10}{7} \Leftrightarrow x = \dfrac{1}{2}$
Giải thích các bước giải:
$\begin{array}{l}P(x) = \dfrac{2x^2 - 2x + 3}{x^2 - x + 2}\\ =\dfrac{2(x^2 -x + 2) - 1}{x^2 - x+ 2}\\ = 2 - \dfrac{1}{x^2 - x + 2}\\ \text{Ta có:}\\ x^2 - x + 2 = x^2 - 2. \dfrac{1}{2}.x + \dfrac{1}{4} + \dfrac{7}{4} = \left(x - \dfrac{1}{2}\right)^2 + \dfrac{7}{4}\\ Do\,\,\,\left(x - \dfrac{1}{2}\right)^2 \geq 0, \forall x\\ nên\,\,\,\left(x - \dfrac{1}{2}\right)^2 + \dfrac{7}{4} \geq \dfrac{7}{4}\\ \Leftrightarrow \dfrac{1}{\left(x - \dfrac{1}{2}\right)^2 + \dfrac{7}{4}}\leq \dfrac{4}{7}\\ \Leftrightarrow -\dfrac{1}{\left(x - \dfrac{1}{2}\right)^2 + \dfrac{7}{4}}\geq -\dfrac{4}{7}\\ \Leftrightarrow 2 -\dfrac{1}{\left(x - \dfrac{1}{2}\right)^2 + \dfrac{7}{4}}\geq 2 -\dfrac{4}{7} = \dfrac{10}{7}\\ Hay\,\,\,P(x) \geq \dfrac{10}{7}\\ \text{Dấu = xảy ra}\, \Leftrightarrow x - \dfrac{1}{2} = 0 \Leftrightarrow x = \dfrac{1}{2}\\ Vậy\,\,\,\,minP = \dfrac{10}{7} \Leftrightarrow x = \dfrac{1}{2}\end{array}$
