$(x+3)^6.(2x+1)^4$
$=\sum\limits_{k=0}^6.C_6^k.x^{6-k}.3^k.\sum\limits_{l=0}^4.C_4^l.2^{4-l}.x^{4-l}$
$\Rightarrow 6-k+4-l=8$
$\Leftrightarrow k+l=2$
$\Rightarrow (k;l)=(0;2), (1;1), (2;0)$
Số cần tìm có thể là:
$C_6^0.C_4^2.3^0.2^2.x^8=24x^8$
$C_6^1.C_4^1.3^1.2^3.x^8=576x^8$
$C_6^2.C_4^0.3^2.2^4.x^8=2160x^8$