Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
B = {\left( {x + 2} \right)^2} + {\left( {x - 2} \right)^2} - 2.\left( {x + 2} \right)\left( {x - 2} \right)\\
= {\left( {x + 2} \right)^2} - 2.\left( {x + 2} \right).\left( {x - 2} \right) + {\left( {x - 2} \right)^2}\\
= {\left[ {\left( {x + 2} \right) - \left( {x - 2} \right)} \right]^2}\\
= {4^2}\\
= 16,\,\,\,\forall x\\
\Rightarrow x = - 4 \Rightarrow B = 16\\
2,\\
4{x^2} - 4x + 1 = {\left( {2x} \right)^2} - 2.2x.1 + {1^2} = {\left( {2x - 1} \right)^2}\\
3,\\
A = \dfrac{3}{{{x^2} + 2x + 3}}\\
{x^2} + 2x + 3 = \left( {{x^2} + 2x + 1} \right) + 2 = {\left( {x + 1} \right)^2} + 2 \ge 2,\,\,\,\forall x\\
\Rightarrow \dfrac{3}{{{x^2} + 2x + 3}} \le \dfrac{3}{2},\,\,\,\forall x\\
\Rightarrow A \le \dfrac{3}{2},\,\,\,\forall x\\
\Rightarrow {A_{\max }} = \dfrac{3}{2} \Leftrightarrow {\left( {x + 1} \right)^2} = 0 \Leftrightarrow x = - 1
\end{array}\)
