Giải thích các bước giải:
+) Tìm GTNN:
Ta có:
$\begin{array}{l}
B + \dfrac{1}{2} = \dfrac{{2x + 3}}{{{x^2} + 2x + 3}} + \dfrac{1}{2}\\
= \dfrac{{{x^2} + 6x + 9}}{{2\left( {{x^2} + 2x + 3} \right)}}\\
= \dfrac{{{{\left( {x + 3} \right)}^2}}}{{2\left( {{{\left( {x + 1} \right)}^2} + 2} \right)}}\\
\text{Do}\left\{ \begin{array}{l}
{\left( {x + 3} \right)^2} \ge 0,\forall x\\
{\left( {x + 1} \right)^2} + 2 > 0,\forall x
\end{array} \right.\\
\Rightarrow \dfrac{{{{\left( {x + 3} \right)}^2}}}{{{{\left( {x + 1} \right)}^2} + 2}} \ge 0,\forall x\\
\Rightarrow B + \dfrac{1}{2} \ge 0\\
\Rightarrow B \ge \dfrac{{ - 1}}{2}\\
\text{Dấu bằng xảy ra} \Leftrightarrow {\left( {x + 3} \right)^2} = 0 \Leftrightarrow x = - 3\\
\text{Vậy}MinB = \dfrac{{ - 1}}{2} \Leftrightarrow x = - 3
\end{array}$
+) Tìm GTLN:
Ta có:
$\begin{array}{l}
B = \dfrac{{2x + 3}}{{{x^2} + 2x + 3}}\\
B - 1 = \dfrac{{2x + 3}}{{{x^2} + 2x + 3}} - 1\\
= \dfrac{{ - {x^2}}}{{{x^2} + 2x + 3}}\\
= \dfrac{{ - {x^2}}}{{{{\left( {x + 1} \right)}^2} + 2}}\\
\text{Do}\left\{ \begin{array}{l}
{x^2} \ge 0,\forall x\\
{\left( {x + 1} \right)^2} + 2 \ge 2 > 0,\forall x
\end{array} \right.\\
\Rightarrow \dfrac{{ - {x^2}}}{{{{\left( {x + 1} \right)}^2} + 2}} \le 0\\
\Rightarrow B - 1 \le 0\\
\Rightarrow B \le 1\\
\text{Dấu bằng xảy ra}\Leftrightarrow {x^2} = 0 \Leftrightarrow x = 0\\
\text{Vậy} MaxB = 1 \Leftrightarrow x = 0
\end{array}$
