$P=\dfrac{x^4-x^2}{x^2+5x+6}\\ =\dfrac{x^2(x^2-1)}{(x+2)(x+3)}\\ =\dfrac{x^2(x-1)(x+1)}{(x+2)(x+3)} \le 0\\ \Leftrightarrow \dfrac{(x-1)(x+1)}{(x+2)(x+3)} \le 0( Do \, \, x^2 \ge 0\, \forall\,x)$
Lập bảng xét dấu:
\begin{array}{|c|cccccc|} \hline x&-\infty&-3&&-2&&-1&&1&+\infty\\\hline P&\,\,\,+&\,\,\,\,||&-&\,\,||&+&\,\,\,\,0&-&0&+ \\\hline \end{array}
Dựa vào bảng xét dấu
$\Rightarrow P \le 0 \Leftrightarrow \left[\begin{array}{l} x \ge 1\\2 <x\le -1\\x<-3\end{array} \right.$
