what should be approach?

\hspace{-16}$Given $\bf{A=2010^{4019}}$ and $\bf{B=2009^{2009}\cdot 2011^{2011}}$\\\\\\ Let $\bf{2010 = x\;,}$ Then $\bf{B=(x-1)^{(x-1)}\cdot (x+1)^{(x+1)}}$\\\\\\ So $\bf{B=(x-1)^x\cdot (x+1)^x\cdot \left(\frac{x+1}{x-1}\right)=(x^2-1)^{x}\cdot \left(1+\frac{2}{x-1}\right)}$\\\\\\ Now If $\bf{x=}$ Large positive Quantity. We can write as\\\\\\ $\bf{B = \left(x^2-1\right)^x\cdot \left(1+\frac{2}{x-1}\right)\approx x^{2x}>x^{2x-1}=A}$\\\\\\ So From above Conclusion $\boxed{\boxed{\bf{ B>A}}}$