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\(A=1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2018}+\frac{1}{2019}\)
=>\(A=1+\frac12+\frac13+\cdots+\frac{1}{2018}+\frac{1}{2019}-2\left(\frac12+\frac14+\cdots+\frac{1}{2018}\right)\)
\(=1+\frac12+\cdots+\frac{1}{2019}-1-\frac12-\cdots-\frac{1}{1009}\)
\(=\frac{1}{1010}+\frac{1}{1011}+\cdots+\frac{1}{2019}\)
=B
=>A-B=0
=>\(\left(A-B\right)^{2019}=0\)
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
\(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2018}+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}=N\)
\(\Rightarrow M-N=0\Rightarrow\left(M-N\right)^2=0\)