$\Big(-4x+\dfrac{1}{4x^2}\Big)^{11}$
$=\sum\limits_{k=0}^{11}C_{11}^k.(-4)^{11-k}.x^{11-k}.\dfrac{1}{4^k}.\dfrac{1}{x^{2k}}$
$=\sum\limits_{k=0}^{11}C_{11}^k.(-4)^{11-k}.\dfrac{1}{4^k}.x^{11-3k}$
$\Rightarrow 11-3k=2\Leftrightarrow k=3$
Hệ số số hạng chứa $x^2$:
$C_{11}^3.(-4)^8.\dfrac{1}{4^3}=168960$