Đáp án:
\(Min = \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{{x^2} + 2x + 2}}{{{x^2} + 2x + 3}} = \dfrac{{{x^2} + 2x + 3 - 1}}{{{x^2} + 2x + 3}}\\
= 1 - \dfrac{1}{{{x^2} + 2x + 1 + 2}}\\
= 1 - \dfrac{1}{{{{\left( {x + 1} \right)}^2} + 2}}\\
Do:{\left( {x + 1} \right)^2} \ge 0\forall x\\
\to {\left( {x + 1} \right)^2} + 2 \ge 2\\
\to - \dfrac{1}{{{{\left( {x + 1} \right)}^2} + 2}} \ge - \dfrac{1}{2}\\
\to 1 - \dfrac{1}{{{{\left( {x + 1} \right)}^2} + 2}} \ge \dfrac{1}{2}\\
\to Min = \dfrac{1}{2}\\
\Leftrightarrow x + 1 = 0\\
\Leftrightarrow x = - 1
\end{array}\)
