Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1/1-1/2+1/2-1/3+1/3-...-1/2020+1/2020-1/2021
=1/1+(1/2-1/2)+(1/3-1/3)+...+(1/2020-1/2020)-1/2021
=1/1-1/2021
=1-1/2021
=2020/2021
Học tốt nha!!!
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)
=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)
Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)
=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)
=> 10B < 10A
=> B < A
b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)
Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)
=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> B < A
bài 1:
ssh của A là:
(151-3):2+1=75
A=(151+3)x75:2=5775
đáp số: 5775
Ta có: \(B=1\cdot2\cdot3\cdot\ldots\cdot2020\cdot\left(1+\frac12+\cdots+\frac{1}{2020}\right)\)
\(=1\cdot2\cdot3\cdot\ldots\cdot2020\cdot\left(1+\frac{1}{2020}+\frac12+\frac{1}{2019}\cdots+\frac{1}{1010}+\frac{1}{1011}\right)\)
\(=1\cdot2\cdot3\cdot\ldots\cdot2020\cdot\left(\frac{2021}{1\cdot2020}+\frac{2021}{2\cdot2019}+\cdots+\frac{2021}{1010\cdot1011}\right)\)
\(=1\cdot2\cdot3\cdot\ldots\cdot2020\cdot2021\cdot\left(\frac{1}{1\cdot2020}+\frac{1}{2\cdot2019}+\cdots+\frac{1}{1010\cdot1011}\right)\)
=>B⋮2021