Hệ số của $x^k$ trong khai triển là: $a_k=C_{10}^k\left(\dfrac{2}{2013}\right)^k\left(\dfrac{2010}{2012}\right)^{10-k}$
Ta sẽ chứng minh: $a_k>a_{k+1},\forall k$
$\Leftrightarrow C_{10}^k\left(\dfrac{2}{2013}\right)^k\left(\dfrac{2010}{2012}\right)^{10-k}>C_{10}^{k+1}\left(\dfrac{2}{2013}\right)^{k+1}\left(\dfrac{2010}{2012}\right)^{9-k}$
$\Leftrightarrow \dfrac{10!}{k!(10-k)!}\left(\dfrac{2}{2013}\right)^k\left(\dfrac{2010}{2012}\right)^{10-k}>\dfrac{10!}{(k+1)!(9-k)!}\left(\dfrac{2}{2013}\right)^{k+1}\left(\dfrac{2010}{2012}\right)^{9-k}$
$\Leftrightarrow \dfrac{2010}{2012}(k+1)>\dfrac{2}{2013}(10-k)$, luôn đúng.
Vậy hệ só lớn nhất trong khai triển là: $a_0=\left(\dfrac{2010}{2012}\right)^{10}$