a,
$(2x^2-1)^6$
$=\sum\limits_{k=0}^6.C_6^k.2^{6-k}.x^{12-2k}.(-1)^k$
$\Rightarrow 12-2k=4\Leftrightarrow k=4$
Vậy hệ số là: $C_6^4.2^2=60$
b,
$\Big(\dfrac{x}{3}+\dfrac{4}{x}\Big)^8$
$=\sum\limits_{k=0}^8.C_8^k.\dfrac{1}{3^{8-k}}.x^{8-k}.4^k.\dfrac{1}{x^k}$
$=\sum\limits_{k=0}^8.C_8^k.\dfrac{4^k}{3^{8-k}}.x^{8-2k}$
$\Rightarrow 8-2k=4\Leftrightarrow k=2$
Vậy hệ số là: $C_8^2.\dfrac{4^2}{3^6}=\dfrac{448}{729}$