Ta có: ${{\left( x-1 \right)}^{2018}} \ge 0\forall x$, $\left| y+1 \right| \ge 0\forall y$
Khi đó ${{\left( x-1 \right)}^{2018}}+\left| y+1 \right| \ge 0\forall x;y$
Để ${{\left( x-1 \right)}^{2018}}+\left| y+1 \right|=0$ thì $\left\{ \begin{align}
& {{\left( x-1 \right)}^{2018}}=0 \\
& \left| y+1 \right|=0 \\
\end{align} \right.$$\Rightarrow \left\{ \begin{align}
& x-1=0 \\
& y+1=0 \\
\end{align} \right.$$\Rightarrow \left\{ \begin{align}
& x=1 \\
& y=-2 \\
\end{align} \right.$
Khi đó $P=\frac{{{1}^{2019}}.{{\left( -1 \right)}^{2020}}}{{{\left( 2.1-1 \right)}^{2019+2020}}}=\frac{1.1}{{{1}^{4039}}}=\frac{1}{1}=1$