Gọi \(d=\left(a,b\right)\)(d>0)
\(\Rightarrow\hept{\begin{cases}a=dx\\b=dy\end{cases}}\)(x,y)=1
Ta có \(a^2+b^2=d^2\left(x^2+y^2\right)⋮ab=d^2xy\)
\(\Rightarrow x^2+y^2⋮xy\)
Vì (x,y)=1 nên \(\hept{\begin{cases}x^2+y^2⋮x\\x^2+y^2⋮y\end{cases}\Rightarrow}x=y=1\)
do đó \(A=\frac{d^{2018}\left(x^{2018}+y^{2018}\right)}{d^{2018}.x^{1009}.y^{1008}}=2\)