Ta có: \({x^2} + 2x + 3\) \( = \left( {{x^2} + 2x + 1} \right) + 2\) \( = {\left( {x + 1} \right)^2} + 2 \ge 2\)
\( \Rightarrow \dfrac{4}{{{x^2} + 2x + 3}} \le \dfrac{4}{2} = 2\) \( \Rightarrow y \le 2\).
Dấu “=” xảy ra khi \(x + 1 = 0 \Leftrightarrow x = - 1\).
Vậy \(\max y = 2\) khi \(x = - 1\).
Chọn B.
