Ta có: \(x^2+5xy+4y^2=xy-5\left|y-3\right|\)
=>\(x^2+4xy+4y^2=-5\left|y-3\right|\)
=>\(\left(x+2y\right)^2+5\left|y-3\right|=0\)
=>\(\begin{cases}y-3=0\\ x+2y=0\end{cases}\Rightarrow\begin{cases}y=3\\ x=-2y=-6\end{cases}\)
\(H=\left(x+5\right)^{2020}\cdot\left(y-2\right)^{2021}+xy\)
\(=\left(-6+5\right)^{2020}\cdot\left(3-2\right)^{2021}+3\cdot\left(-6\right)\)
=1-18
=-17

