Question

Các bn giải giúp mìn từ câu a đến h nkaaaa

Answer 1

a: \(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{999}-\dfrac{1}{1000}=\dfrac{999}{1000}\)

b: \(B=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{999}-\dfrac{1}{1000}\right)\)

\(=2\cdot\dfrac{999}{1000}=\dfrac{999}{500}\)

Answer 2

a.

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(A=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)

f.

\(F=-\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2022}+\dfrac{1}{2021}-....-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)

\(F=-\dfrac{1}{2023}+1=\dfrac{2022}{2023}\)