Giải thích các bước giải:
Ta có:
$C = \dfrac{{{x^2}}}{{{x^2} + x + 1}}$
Mà:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} \ge 0,\forall x \ne - 1\\
{x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0,\forall x \ne - 1
\end{array} \right.\\
\Rightarrow \dfrac{{{x^2}}}{{{x^2} + x + 1}} \ge 0,\forall x \ne - 1\\
\Rightarrow C \ge 0\left( 1 \right)
\end{array}$
Lại có:
$\begin{array}{l}
C - \dfrac{4}{3} = \dfrac{{{x^2}}}{{{x^2} + x + 1}} - \dfrac{4}{3}\\
= \dfrac{{ - {x^2} - 4x - 4}}{{{x^2} + x + 1}}\\
= \dfrac{{ - {{\left( {x + 2} \right)}^2}}}{{{x^2} + x + 1}}
\end{array}$
Mặt khác:
$\begin{array}{l}
\left\{ \begin{array}{l}
- {\left( {x + 2} \right)^2} \le 0,\forall x \ne - 1\\
{x^2} + x + 1 > 0,\forall x \ne - 1
\end{array} \right.\\
\Rightarrow \dfrac{{ - {{\left( {x + 2} \right)}^2}}}{{{x^2} + x + 1}} \le 0,\forall x \ne - 1\\
\Rightarrow C - \dfrac{4}{3} \le 0\\
\Rightarrow C \le \dfrac{4}{3}\left( 2 \right)
\end{array}$
Từ $\left( 1 \right),\left( 2 \right) \Rightarrow 0 \le C \le \dfrac{4}{3}$
Mà $C \in Z \Leftrightarrow \left[ \begin{array}{l}
C = 0\\
C = 1
\end{array} \right.$
$\begin{array}{l}
+ )TH1:C = 0\\
\Leftrightarrow {x^2} = 0\\
\Leftrightarrow x = 0\left( {tm} \right)\\
+ )TH2:C = 1\\
\Leftrightarrow \dfrac{{{x^2}}}{{{x^2} + x + 1}} = 1\\
\Leftrightarrow {x^2} + x + 1 = {x^2}\\
\Leftrightarrow x + 1 = 0\\
\Leftrightarrow x = - 1\left( l \right)
\end{array}$
Vậy $x=0$ thỏa mãn đề.
