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\(\Rightarrow2019\left|x-1\right|+2020\left|y-2\right|+2021\left|y-3\right|+2022\left|y-4\right|=2020+2022\)
\(\Rightarrow\hept{\begin{cases}\left|y-2\right|=1\\\left|x-1\right|=0\\\left|y-4\right|=1\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}}\)
Ta có: \(A-B\)
\(=1011\left(1+\frac13+\frac15+\cdots+\frac{1}{2019}\right)-1010\left(\frac12+\frac14+\cdots+\frac{1}{2020}\right)\)
\(=1010\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}\right)+\left(1+\frac13+\cdots+\frac{1}{2019}\right)\)
Vì \(1-\frac12>0;\frac13-\frac14>0;\ldots;\frac{1}{2019}-\frac{1}{2020}>0\)
nên \(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}>0\)
=>\(1010\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}\right)>0\)
=>A-B>0
=>A>B
\(\dfrac{2^{4042}\cdot3^{2020}}{6^{2019}\cdot4^{1011}}=\dfrac{2^{4042}\cdot3^{2020}}{3^{2019}\cdot2^{2019}\cdot2^{2022}}\)
\(=3\cdot\dfrac{2^{2042}}{2^{4041}}=3\cdot2=6\)