Tìm gtln gtnn B=$\frac{2x+3}{x^{2} +2x+3}$

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1. 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:

   $\begin{array}{l}
   B = \dfrac{{2x + 3}}{{{x^2} + 2x + 3}}\\
   B – 1 = \dfrac{{2x + 3}}{{{x^2} + 2x + 3}} – 1\\
    = \dfrac{{ – {x^2}}}{{{x^2} + 2x + 3}}\\
    = \dfrac{{ – {x^2}}}{{{{\left( {x + 1} \right)}^2} + 2}}\\
   \text{Do}\left\{ \begin{array}{l}
   {x^2} \ge 0,\forall x\\
   {\left( {x + 1} \right)^2} + 2 \ge 2 > 0,\forall x
   \end{array} \right.\\
    \Rightarrow \dfrac{{ – {x^2}}}{{{{\left( {x + 1} \right)}^2} + 2}} \le 0\\
    \Rightarrow B – 1 \le 0\\
    \Rightarrow B \le 1\\
    \text{Dấu bằng xảy ra}\Leftrightarrow {x^2} = 0 \Leftrightarrow x = 0\\
   \text{Vậy} MaxB = 1 \Leftrightarrow x = 0
   \end{array}$

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