Xét $f(x)=\frac{4x}{4x^4+1}=\frac{(2x^2+2x+1)-(2x^2-2x+1)}{(2x^2+2x+1)(2x^2-2x+1)}=\frac{1}{2x^2-2x+1}-\frac{1}{2x^2+2x+1}(*)$Lại chứng minh đc $\frac{1}{2x^2+2x+1}=\frac{1}{2(x+1)^2-2(x+1)+1}$
Nên $(*)\Leftrightarrow f(x)=\frac{1}{2x^2-2x+1}-\frac{1}{2(x+1)^2-2(x+1)+1}$
Vậy :$A=f(1)+f(2)+...+f(n)=\frac{1}{2.1^2-2.1+1}-\frac{1}{2(n+1)^2-2(n+1)+1}$
$=1-\frac{1}{2n^2+2n+1}=\boxed{\frac{2n^2+2n}{2n^2+2n+1}}$