Đáp án:
1) $\max\left(\dfrac{1}{x^2 + x + 1}\right) = \dfrac43 \Leftrightarrow x = -\dfrac12$
2) $\max \left(\dfrac{3}{4x^2 -4x + 5}\right) = \dfrac34 \Leftrightarrow x = \dfrac12$
Giải thích các bước giải:
$\begin{array}{l}1)\quad A = \dfrac{1}{x^2 + x + 1}\\ \to A = \dfrac{1}{x^2 + 2\cdot\dfrac12x + \dfrac14 + \dfrac34}\\ \to A = \dfrac{1}{\left(x + \dfrac12\right)^2 + \dfrac34}\\ \text{Ta có:}\\ \quad \left(x + \dfrac12\right)^2\geq 0\quad \forall x\\ \to \left(x + \dfrac12\right)^2 + \dfrac34 \geq \dfrac34\\ \to \dfrac{1}{\left(x + \dfrac12\right)^2+\dfrac34} \leq \dfrac43\\ \to A \leq \dfrac43\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \left(x + \dfrac12\right)^2 = 0\Leftrightarrow x = -\dfrac12\\ Vậy \,\,\max\left(\dfrac{1}{x^2 + x + 1}\right) = \dfrac43 \Leftrightarrow x = -\dfrac12\\ 2)\quad B = \dfrac{3}{4x^2 -4x + 5}\\ \to B = \dfrac{3}{4x^2 - 4x + 1 + 4}\\ \to B = \dfrac{3}{(2x - 1)^2 +4}\\ \text{Ta có:}\\ \quad (2x -1)^2 \geq 0\quad \forall x\\ \to (2x-1)^2 + 4 \geq 4\\ \to \dfrac{3}{(2x-1)^2 + 4} \leq \dfrac{3}{4}\\ \to B \leq \dfrac34\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow (2x - 1)^2 = 0 \Leftrightarrow x = \dfrac12\\ Vậy\,\,\max \left(\dfrac{3}{4x^2 -4x + 5}\right) = \dfrac34 \Leftrightarrow x = \dfrac12 \end{array}$
