$\left \{ {{x^3+2y^2-4y+3=0} \atop {x^2+x^2y^2-2y=0}} \right.$ $\leftrightarrow \left \{ {{x^3+1+2(y-1)^2=0} \atop {x^2(y^2+1)=2y}} \right.$ $\leftrightarrow \left \{ {{2(y-1)^2=-1-x^3 \ (1)} \atop {x^2=\frac{2y}{y^2+1}}} \right.$
Từ (1) suy ra $-1-x^3\geq 0$ $\leftrightarrow -1\geq x^3$ $\leftrightarrow -1\geq x$ $\rightarrow x^2 \geq 1$ (2)
Có: $(y-1)^2\geq 0$ $\leftrightarrow y^2-2y+1\geq 0$ $\leftrightarrow y^2+1 \geq 2y$
$\leftrightarrow 1 \geq \frac{2y}{y^2+1}=x^2$ (3)
Từ (2) và (3) suy ra : $x^2=1$ xảy ra khi $\left \{ {{x=-1} \atop {y=1}} \right.$
Khi đó: $P=x^{2020}+y^{2020}=(-1)^{2020}+1^{2020}=2$
