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$A=\frac{2019^{2019}}{2019^{2018}}=2019>2018=\frac{2018^{2019}}{2018^{2018}}>\frac{2018^{2019}}{2018^{2018}+2018}=B$ (Because the denominator, we get the last inequality.) Hence, $A>B$

$\frac{1}{A}=$ $\frac{1}{2019}$

$\frac{1}{B}=$ $\frac{2018^{2018}}{2018^{2019}}+\frac{2018}{2018^{2019}}=$ $\frac{1}{2018}+\frac{1}{2018^{2018}}>$ $\frac{1}{2018}>$ $\frac{1}{2019}=$ $\frac{1}{A}$

So, $\frac{1}{B}>\frac{1}{A}$ or,

$\boxed{A>B}$