Có: \(\left|2022-2x+y\right|\ge0\forall x,y\)

       \(\left(x-y-2021\right)^2\ge0\forall x,y\)

\(\Rightarrow\left|2022-2x+y\right|+\left(x-y-2021\right)^2\ge0\forall x,y\)

Mặt khác: \(\left|2022-2x+y\right|+\left(x-y-2021\right)^2=0\)

nên \(\left\{{}\begin{matrix}2022-2x+y=0\\x-y-2021=0\end{matrix}\right.\)           \(\Rightarrow\left\{{}\begin{matrix}-2x+y=-2022\\x-y=2021\end{matrix}\right.\)

\(\Rightarrow-2x+y+x-y=-2022+2021\)

\(\Rightarrow-x=-1\Leftrightarrow x=1\)

Khi đó: \(1-y=2021\) \(\Leftrightarrow y=-2020\)

\(\Rightarrow x+y=1-2020=-2019\)

|2022-2x+y|+(x-y-2021)^2=0

=>2022-2x+y=0 và x-y-2021=0

=>x-y=2021 và 2x-y=2022

=>x=1 và y=-2020

kết quả là -2023 nhé mọi ng mình biết kqua nhm k biết trình bày =))

Cứu với ;-;

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2022}\)

\(\Rightarrow\dfrac{yz+zx+xy}{xyz}=\dfrac{1}{x+y+z}\)

\(\Rightarrow\left(yz+zx+xy\right)\left(x+y+z\right)=xyz\)

\(\Rightarrow xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+3xyz-xyz=0\)

\(\Rightarrow xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Rightarrow x=-y\) hoặc \(y=-z\) hoặc \(z=-x\).

-Đến đây thôi bạn, câu hỏi sai rồi ạ.

\(a,\left\{{}\begin{matrix}\left|x-3y\right|\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y=-12\\y=-4\end{matrix}\right.\)

\(b,Sửa:\left|x-y-5\right|+\left(y+3\right)^2=0\\ \left\{{}\begin{matrix}\left|x-y-5\right|\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y-5=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=2\\y=-3\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}\left|x+y-1\right|\ge0\\\left(y-2\right)^4\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

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

\(d,\left\{{}\begin{matrix}\left|x+3y-1\right|\ge0\\3\left|y+2\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+3y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-3y=7\\y=-2\end{matrix}\right.\)

\(e,Sửa:\left|2021-x\right|+\left|2y-2022\right|=0\\ \left\{{}\begin{matrix}\left|2021-x\right|\ge0\\\left|2y-2022\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2021-x=0\\2y-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\y=1011\end{matrix}\right.\)

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

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

\(\left|x-2\right|+\left|y-1\right|+\left(x+y-z-2\right)^{2022}=0\)

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

\(A=5\cdot2^2\cdot1^{2020}\cdot1^{2021}=20\)