Đáp án:
$MinA = \dfrac{{ - 1}}{3} \Leftrightarrow x = - 2$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
A = \dfrac{{x + 1}}{{{x^2} + x + 1}}\\
\Rightarrow A + \dfrac{1}{3} = \dfrac{{x + 1}}{{{x^2} + x + 1}} + \dfrac{1}{3}\\
= \dfrac{{3\left( {x + 1} \right) + {x^2} + x + 1}}{{3\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{{x^2} + 4x + 4}}{{3\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{{{\left( {x + 2} \right)}^2}}}{{3\left( {{x^2} + x + 1} \right)}}
\end{array}$
Mà:
$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} \ge 0,\forall x\\
{x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0,\forall x
\end{array} \right.\\
\Rightarrow \dfrac{{{{\left( {x + 2} \right)}^2}}}{{3\left( {{x^2} + x + 1} \right)}} \ge 0,\forall x\\
\Rightarrow A + \dfrac{1}{3} \ge 0,\forall x\\
\Rightarrow A \ge \dfrac{{ - 1}}{3}
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow {\left( {x + 2} \right)^2} = 0 \Leftrightarrow x + 2 = 0 \Leftrightarrow x = - 2$
Vậy $MinA = \dfrac{{ - 1}}{3} \Leftrightarrow x = - 2$
