Đáp án:
$\min\left[\dfrac{2(x^2+ x+1)}{x^2+ 1}\right] = 1 \Leftrightarrow x = -1$
$\max\left[\dfrac{2(x^2+ x+1)}{x^2+ 1}\right] = 3 \Leftrightarrow x = 1$
Giải thích các bước giải:
$\begin{array}{l}Đặt \,\,P =\dfrac{2(x^2+ x+1)}{x^2+ 1}\\ +) \quad P - 1 = \dfrac{2(x^2 + x + 1)}{x^2 + 1} - 1\\ \to P - 1 = \dfrac{2(x^2 + x +1) -(x^2 + 1)}{x^2 + 1}\\ \to P - 1 = \dfrac{x^2 + 2x + 1}{x^2 + 1}\\ \to P - 1 = \dfrac{(x+1)^2}{x^2+1} \geq 0\\ \to P - 1 \geq 0\\ \to P \geq 1\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x + 1 =0\Leftrightarrow x = -1\\ \to \min P = 1 \Leftrightarrow x = -1\\ +) \quad P -3 =\dfrac{2(x^2 + x + 1)}{x^2 + 1} - 3\\ \to P - 3= \dfrac{2(x^2 + x +1) -3(x^2 + 1)}{x^2 + 1}\\ \to P -3 = \dfrac{-x^2 + 2x - 1}{x^2 + 1}\\ \to P - 3 = -\dfrac{(x-1)^2}{x^2 +1} \leq 0\\ \to P - 3 \leq 0\\ \to P \leq 3\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x - 1 =0 \Leftrightarrow x = 1\\ \to \max P = 3 \Leftrightarrow x = 1 \end{array}$
