Đặt $x=2003$
[tex]\Rightarrow E=\frac{\left [ x^2(x+10)+31(x+1)-1 \right ].\left [ x(x+5)+4 \right ]}{(x+1)(x+2)(x+3)(x+4)(x+5)}[/tex]
Ta có: [tex]x^2(x+10)+31(x+1)-1=x^3+10x^2+31x+30=(x+2)(x+3)(x+5) \\ x(x+5)+4=(x+1)(x+4)[/tex]
Thay vào ta được: $E=1$