Ta thấy : \(\left|x-2021\right|\ge0\forall x,\left|y-2022\right|\ge0\forall y\\ =>\left|x-2021\right|+\left|y-2022\right|\ge0\)

Mà theo đề : \(\left|x-2021\right|+\left|y-2022\right|\le0\)

=> \(\left\{{}\begin{matrix}x-2021=0\\y-2022=0\end{matrix}\right.=>\left(x;y\right)=\left(2021;2022\right)\)

tick mk nha các bn ơi

xạo vừa vừa thôi mấy mắm ơi, chtt đâu có đâu

\(\left(\sqrt{x-1}+\sqrt{3-x}\right)^2\le\left(1^2+1^2\right)\left(x-1+3-x\right)=4\\ \Leftrightarrow\sqrt{x-1}+\sqrt{3-x}\le2\\ y^2+2\sqrt{2020}y+2022=\left(y^2+2y\sqrt{2020}+2020\right)+2\\ =\left(y+\sqrt{2020}\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=3-x\\y+\sqrt{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\sqrt{2020}\end{matrix}\right.\)

Vậy ...

ĐKXĐ: \(3\ge x\ge1\)

Áp dụng BĐT Bunhiacopski:

\(1\sqrt{x-1}+1\sqrt{3-x}\le\sqrt{\left(1^2+1^2\right)\left(x-1+3-x\right)}=\sqrt{2.2}=2\)

Mặt khác: \(y^2+2\sqrt{2020}y+2022=\left(y+\sqrt{2020}\right)^2+2\ge2\)

Nên để thõa mãn yêu cầu bài toán thì

\(\left\{{}\begin{matrix}\sqrt{x-1}=\sqrt{3-x}\\y+\sqrt{2020}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(tm\right)\\y=-\sqrt{2020}\end{matrix}\right.\)

\(P=\dfrac{1}{2021}\left(\dfrac{2021^2}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{2021}.\dfrac{\left(2021+1\right)^2}{x+y}=\dfrac{1}{2021}.\dfrac{2022^2}{\dfrac{2022}{2021}}=2022\)

\(P_{min}=2022\) khi \(\left(x;y\right)=\left(1;\dfrac{1}{2021}\right)\)

\(\Rightarrow2019\left|x-1\right|+2020\left|y-2\right|+2021\left|y-3\right|+2022\left|y-4\right|=2020+2022\)

\(\Rightarrow\hept{\begin{cases}\left|y-2\right|=1\\\left|x-1\right|=0\\\left|y-4\right|=1\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}}\)

ta có :

2022 mà bạn