cho mình sửa một chút 1998->1988

\(\frac{2003\times14+1998+2001\times2002}{2002+2002\times503+504\times2002}=\frac{\left(2002+1\right)\times14+1998+2001\times2002}{2002\left(1+503+504\right)}=\frac{2002\times14+14+1998+2001\times2002}{2002\times1008}\)=\(\frac{2002\times14+2002\times2001+2002}{2002\times1008}=\frac{2002\left(14+2001+1\right)}{2002\times1008}=\frac{2002\times2016}{2002\times1008}\)=\(\frac{2016}{1008}=\frac{2}{1}=2\)