Ta có : \(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)

\(=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2019}=B\)

\(\Rightarrow A-B-1=-1\)

\(\Rightarrow\left(A-B-1\right)^{2019}=-1\)

Ơ !!! Bài này giống bài 5 môn Toán thi cuối học kỳ 2 trường mình nè !!!

Kết quả là -1 thì phải !!!

ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(A=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(A=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}+\frac{1}{2018}\)

\(\Rightarrow A=B\left(=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}+\frac{1}{2018}\right)\)

\(\Rightarrow\frac{A}{B^{2018}}=\frac{A}{A.B^{2017}}=\frac{1}{B^{2017}}\)

=> \(\frac{A}{B^{2018}}=\frac{1}{\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}+\frac{1}{2018}\right)^{2017}}\)

A=(1+1/3+...+1/2019)-(1/2+1/4+...+1/2018)

A=(1+1/3+...+1/2019)+(1/2+1/4+...+1/2018)-(1/2+1/4+...+1/2018).2

A=(1+1/2+1/3+1/4+...+1/2019)-(1+1/2+...+1/1009)

A=1/1010+1/1011+...+1/2019

=) A=B

=) (A-B-1)^2019=-1