$\dfrac{x^2+x-12}{x^2-36}=\dfrac{(x-3)(x+4)}{(x-6)(x+6)}$
Lập bảng xét dấu
\begin{array}{|c|cccccccccccccc|} \hline x&-\infty&&-6&&-4&&3&&6&&+\infty\\\hline x^2+x-12&&+&|&+&0&-&0&+&|&+& \\\hline x^2-36&&+&0&-&|&-&|&-&0&+& \\\hline\dfrac{x^2+x-12}{x^2-36}&&+&\Bigg|\Bigg|&-&0&+&0&-&\Bigg|\Bigg|&+&\\\hline\end{array}