Ta có: \(\left(4x-7y+2020\right)^{2012}>=0\forall x,y\)
\(\left|x-3y\right|^{2623}>=0\forall x,y\)
Do đó: \(\left(4x-7y+2020\right)^{2012}+\left|x-3y\right|^{2623}>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}4x-7y+2020=0\\x-3y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3y\\4\cdot3y-7y+2020=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3y\\5y=-2020\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-404\\x=3\cdot\left(-404\right)=-1212\end{matrix}\right.\)
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