Cho khai triển $T={{\left( 1+x-{{x}^{2018}}...

$\begin{aligned}
& T={{\left( 1+x-{{x}^{2018}} \right)}^{2019}}+{{\left( 1-x+{{x}^{2019}} \right)}^{2018}} \\
& \ \ \ ={{\sum\limits_{k=0}^{2019}{C_{2019}^{k}{{1}^{2019-k}}\left( x-{{x}^{2018}} \right)}}^{k}}+\sum\limits_{m=0}^{2018}{C_{2018}^{m}{{1}^{2018-m}}{{\left( -x+{{x}^{2019}} \right)}^{m}}} \\
& \ \ \ =\sum\limits_{k=0}^{2019}{C_{2019}^{k}}\sum\limits_{h=0}^{k}{C_{k}^{h}{{x}^{k-h}}{{\left( -1 \right)}^{h}}{{\left( {{x}^{2018}} \right)}^{h}}}+\sum\limits_{m=0}^{2018}{C_{2018}^{m}}\sum\limits_{n=0}^{m}{C_{m}^{n}{{\left( -1 \right)}^{m-n}}{{x}^{m-n}}{{\left( {{x}^{2019}} \right)}^{n}}} \\
& \ \ \ =\sum\limits_{k=0}^{2019}{C_{2019}^{k}}\sum\limits_{h=0}^{k}{C_{k}^{h}{{\left( -1 \right)}^{h}}{{x}^{2017h+k}}}+\sum\limits_{m=0}^{2018}{C_{2018}^{m}}\sum\limits_{n=0}^{m}{C_{m}^{n}{{\left( -1 \right)}^{m-n}}{{x}^{2018n+m}}} \\
\end{aligned}$
Với $0\le h\le k\le 2019;\ 0\le n\le m\le 2018;\ h,k,m,n\in \mathbb{N}.$
Theo đề bài: $\left\{ \begin{aligned}
& 2017h+k=1 \\
& 2018n+m=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& h=0,\ k=1 \\
& n=0,\ m=1 \\
\end{aligned} \right..$
Hệ số của số hạng chứa x trong khai triển của T là $C_{2019}^{1}C_{1}^{0}{{\left( -1 \right)}^{0}}+C_{2018}^{1}C_{1}^{0}{{\left( -1 \right)}^{1-0}}=1.$