\(\left|\left(x-2\right)^{2019}\right|+\left(y-1\right)^{2020}\le0\)

Ta có:

\(\left\{{}\begin{matrix}\left|\left(x-2\right)^{2019}\right|\ge0\\\left(y-1\right)^{2020}\ge0\end{matrix}\right.\forall x,y.\)

\(\Rightarrow\left|\left(x-2\right)^{2019}\right|+\left(y-1\right)^{2020}\ge0\) \(\forall x,y.\)

Mà \(\left|\left(x-2\right)^{2019}\right|+\left(y-1\right)^{2020}\le0.\)

\(\Rightarrow\left|\left(x-2\right)^{2019}\right|+\left(y-1\right)^{2020}=0\)

\(\Rightarrow\left(x-2\right)^{2019}+\left(y-1\right)^{2020}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^{2019}=0\\\left(y-1\right)^{2020}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0+2\\y=0+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{2;1\right\}.\)

Chúc bạn học tốt!

\(x=\frac{2019^{2020}+1}{2019^{2019}+1}>\frac{2019^{2020}+1+2018}{2019^{2019}+1+2018}=\frac{2019^{2020}+2019}{2019^{2019}+2019}=\frac{2019\left(2019^{2019}+1\right)}{2019\left(2019^{2018}+1\right)}=\frac{2019^{2019}+1}{2019^{2018}+1}\)(1)

\(y=\frac{2019^{2019}+2020}{2019^{2018}+2020}< \frac{2019^{2019}+2020-2019}{2019^{2018}+2020-2019}=\frac{2019^{2019}+1}{2019^{2018}+1}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow x>y\)

x-y-z=0

=>x=y+z và y=x-z và z=x-y

B=(1-z/x)(1-x/y)(1+y/z)+2023

\(=\dfrac{x-z}{x}\cdot\dfrac{y-x}{y}\cdot\dfrac{y+z}{z}+2023\)

\(=\dfrac{y}{x}\cdot\dfrac{-z}{y}\cdot\dfrac{x}{z}+2023=2023-1=2022\)

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