Đáp án:
$\begin{array}{l}
Dkxd:x \ne 2;x \ne - 2;x \ne - 3\\
A = \dfrac{{x + 1}}{{x + 2}};B = \dfrac{{{x^3} + 1}}{{2 + x}}\\
A:B \le \dfrac{1}{3}\\
\Rightarrow \dfrac{{x + 1}}{{x + 2}}:\dfrac{{{x^3} + 1}}{{2 + x}} \le \dfrac{1}{3}\\
\Rightarrow \dfrac{{x + 1}}{{x + 2}}.\dfrac{{x + 2}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} \le \dfrac{1}{3}\\
\Rightarrow \dfrac{1}{{{x^2} - x + 1}} \le \dfrac{1}{3}\\
\Rightarrow {x^2} - x + 1 \ge 3\\
\Rightarrow {x^2} - x - 2 \ge 0\\
\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) \ge 0\\
\Rightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le - 1
\end{array} \right.\\
Do:x \ne 2;x \ne - 2;x \ne - 3\\
\Rightarrow x > 2/x \le - 1;x \ne - 2;x \ne - 3
\end{array}$
