Ta có: \(\left|x-2020\right|\ge0\forall x\)

\(\left|y-2021\right|\ge0\forall y\)

Do đó: \(\left|x-2020\right|+\left|y-2021\right|\ge0\forall x,y\)

mà \(\left|x-2020\right|+\left|y-2021\right|=0\)

nên \(\left\{{}\begin{matrix}x-2020=0\\y-2021=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2020\\y=2021\end{matrix}\right.\)

Vậy: (x,y)=(2020;2021)

làm nốt câu này rồi đi ngủ 

\(Q=\frac{|x-2020|+|x-2019|+2019+1}{|x-2019|+|x-2020|+2019}=1+\frac{1}{|x-2020|+|x-2019|+2019}\)

Để Q đạt GTLN thì \(|x-2020|+|x-2019|+2019\)đạt GTNN 

Ta có : \(|x-2020|+|x-2019|+2019=|x-2020|+|2019-x|+2019\)

Sử dụng BĐT /a/ + /b/ >= /a+b/ ta được : 

\(|x-2020|+|2019-x|+2019\ge|x-2020+2019-x|+2019=2020\)

Dấu = xảy ra khi và chỉ khi \(\left(x-2020\right)\left(2019-x\right)\ge0\Leftrightarrow2020\ge x\ge2019\)

Khi đó : \(Q=1+\frac{1}{|x-2020|+|x-2019|+2019}\le1+\frac{1}{2020}=\frac{2021}{2020}\)

Dấu = xảy ra khi và chỉ khi \(2019\le x\le2020\)

Ta có: \(C=\frac{\left|x-2019\right|+2020}{\left|x-2019\right|+2021}\)

\(=\frac{\left|x-2019\right|+2021-1}{\left|x-2019\right|+2021}=1-\frac{1}{\left|x-2019\right|+2021}\)

Ta có: \(\left|x-2019\right|\ge0\forall x\)

=>\(\left|x-2019\right|+2021\ge2021\forall x\)

=>\(\frac{1}{\left|x-2019\right|+2021}\le\frac{1}{2021}\forall x\)

=>\(\frac{-1}{\left|x-2019\right|+2021}\ge\frac{-1}{2021}\forall x\)

=>\(\frac{-1}{\left|x-2019\right|+2021}+1\ge\frac{-1}{2021}+1=\frac{2020}{2021}\forall x\)

Dấu '=' xảy ra khi x-2019=0

=>x=2019