$\Big(2x-\dfrac{1}{x^2}\Big)^{40}$
$=\sum\limits_{k=0}^{40}C_{40}^k.2^{40-k}.x^{40-k}.(-1)^k.\dfrac{1}{x^{2k}}$
$=\sum\limits_{k=0}^{40}C_{40}^k.2^{40-k}.(-1)^k.x^{40-3k}$
$\Rightarrow 40-3k=7\Leftrightarrow k=11$
Vậy hệ số $x^7$ là:
$C_{40}^{11}.2^{40-11}.(-1)$
$=-2^{29}.C_{40}^{11}$