a: \(A=2023-2022+2021-2020+...+3-2+1\)
\(=\left(2023-2022\right)+\left(2021-2020\right)+...+\left(3-2\right)+1\)
\(=1+1+...+1\)
Từ 2 đến 2023 sẽ có: \(\dfrac{2023-2}{1}+1=2023-1=2022\left(số\right)\)
Ta có: 2023-2022=2021-2020=...=3-2=1
=>Sẽ có \(\dfrac{2022}{2}=1011\) cặp số có tổng là 1
=>A=1011*1+1=1012
b: \(B=\dfrac{7}{8}:\left(\dfrac{2}{9}-\dfrac{1}{18}\right)-\dfrac{7}{8}:\left(\dfrac{5}{30}+\dfrac{1}{9}\right)\)
\(=\dfrac{7}{8}:\dfrac{4-1}{18}-\dfrac{7}{8}:\dfrac{15+10}{90}\)
\(=\dfrac{7}{8}\cdot\dfrac{18}{3}-\dfrac{7}{8}\cdot\dfrac{90}{25}\)
\(=\dfrac{7}{8}\left(6-3,6\right)\)
\(=\dfrac{7}{8}\cdot2,4=2,1\)