Giải thích các bước giải:
Đặt $A = \dfrac{{x + 1}}{{{x^2} + x + 1}}$
Ta có:
$\begin{array}{l}
+ )A - 1 = \dfrac{{x + 1}}{{{x^2} + x + 1}} - 1\\
= \dfrac{{ - {x^2}}}{{{x^2} + x + 1}}\\
= \dfrac{{ - {x^2}}}{{{{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}}\\
\le 0,\forall x\\
\Rightarrow A - 1 \le 0,\forall x\\
\Rightarrow A \le 1\left( * \right)\\
+ )A + \dfrac{1}{3} = \dfrac{{x + 1}}{{{x^2} + x + 1}} + \dfrac{1}{3}\\
= \dfrac{{{x^2} + 4x + 4}}{{3\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{{{\left( {x + 2} \right)}^2}}}{{3\left( {{{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}} \right)}}\\
\ge 0,\forall x\\
\Rightarrow A + \dfrac{1}{3} \ge 0,\forall x\\
\Rightarrow A \ge \dfrac{{ - 1}}{3}\left( {**} \right)
\end{array}$
Từ $\left( * \right),\left( {**} \right) \Rightarrow \dfrac{{ - 1}}{3} \le A \le 1$
Để $A \in Z \Leftrightarrow \left[ \begin{array}{l}
A = 0\\
A = 1
\end{array} \right.$
$\begin{array}{l}
+ )TH1:A = 0\\
\Leftrightarrow \dfrac{{x + 1}}{{{x^2} + x + 1}} = 0\\
\Leftrightarrow x + 1 = 0\\
\Leftrightarrow x = - 1\\
+ )TH2:A = 1\\
\Leftrightarrow \dfrac{{x + 1}}{{{x^2} + x + 1}} = 1\\
\Leftrightarrow x + 1 = {x^2} + x + 1\\
\Leftrightarrow {x^2} = 0\\
\Leftrightarrow x = 0
\end{array}$
Vậy $x \in \left\{ { - 1;0} \right\}$ thỏa mãn đề.
