B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022

B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\) 

B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\) 

B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))

Vậy B > C 

tui làm được câu c thui
c) (1-1/2).(1-1/3).(1-1/4).(1-1/5)...(1-1/2022).(1-1/2023)
= 1 2 3 4 2 3 4 5 . . . . . 2021 2022 2022 2023 = 1.2.3.4.5....2021.2022 2.3.4.5....2022.2023 = 1 2023

a: \(\frac{2022\cdot2023-2022}{2021\cdot2022+2022}\)

\(=\frac{2022\left(2023-1\right)}{2022\left(2021+1\right)}\)

\(=\frac{2022\cdot2022}{2022\cdot2022}=1\)

c: \(\left(1-\frac12\right)\left(1-\frac13\right)\cdot\ldots\cdot\left(1-\frac{1}{2022}\right)\left(1-\frac{1}{2023}\right)\)

\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{2021}{2022}\cdot\frac{2022}{2023}\)

\(=\frac{1}{2023}\)

Ta có: \(\frac{2022}{1}+\frac{2021}{2}+\cdots+\frac{1}{2022}\)

\(=\left(\frac{2021}{2}+1\right)+\left(\frac{2020}{3}+1\right)+\cdots+\left(\frac{1}{2022}+1\right)+1\)

\(=\frac{2023}{2}+\frac{2023}{3}+\cdots+\frac{2023}{2023}=2023\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)\)

Ta có: \(\frac{\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)}{\frac{2022}{1}+\frac{2021}{2}+\cdots+\frac{1}{2022}}\)

\(=\frac{\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)}{2023\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)}\)

\(=\frac{1}{2023}\)

Ta có; \(B=1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2022}+\frac{1}{2023}\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2023}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\ldots+\frac{1}{2023}-1-\frac12-\cdots-\frac{1}{1011}=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2023}\)

=C

=>B-C=0

Ta có: C = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/2021.2022.2023

=> C = 1/2. (3-1/1.2.3 + 4-2/2.3.4 + 5-3/3.4.5 + ... + 2023-2021/2021.2022.2023

=> C = 1/2. (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/2021.2022 - 1/2022.2023)

=> C = 1/2. (1/1.2 - 1/2022.2023)

- Phần còn lại bạn tự tính chứ số to quá

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))

vậy x= 2023

Ta có: \(\left(\frac12-1\right)\left(\frac13-1\right)\cdot\ldots\cdot\left(\frac{1}{2023}-1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\ldots\cdot\frac{-2022}{2023}\)

\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{2022}{2023}=\frac{1}{2023}\)

...

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@Bé Bin bạn có biết cách giải không zậy