Question

Answer 1

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: \(B=\dfrac{7}{8}:\left(\dfrac{2}{9}-\dfrac{1}{18}\right)-\dfrac{7}{8}:\left(\dfrac{5}{30}+\dfrac{1}{9}\right)\)

\(=\dfrac{7}{8}:\dfrac{4-1}{18}-\dfrac{7}{8}:\dfrac{15+10}{90}\)

\(=\dfrac{7}{8}\cdot\dfrac{18}{3}-\dfrac{7}{8}\cdot\dfrac{90}{25}\)

\(=\dfrac{7}{8}\left(6-3,6\right)\)

\(=\dfrac{7}{8}\cdot2,4=2,1\)