B=(2020/2+1)+(2019/3+1)+...+(1/2021+1)+1
\(=\dfrac{2022}{2}+\dfrac{2022}{3}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}\)
=>A/B=1/2022
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Các câu hỏi tương tự
cho A=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2022}\)
B=\(\dfrac{2021}{1}+\dfrac{2020}{2}+\dfrac{2019}{3}+...+\dfrac{1}{2021}\)
tính tỉ số \(\dfrac{B}{A}\)
20 tháng 4 2022 lúc 22:36
giúp mk, please :)
\(\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{2017+\dfrac{2016}{6}+\dfrac{2015}{7}+...+\dfrac{1}{2021}}\)
A. \(\dfrac{1}{2020}\)
B. \(\dfrac{1}{2021}\)
C. \(\dfrac{1}{2019}\)
D. \(\dfrac{1}{2022}\)
chọn ra 3 ngừi nhanh nhứt:>>
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}}{\dfrac{2022}{1}+\dfrac{2021}{2}+\dfrac{2020}{3}+...+\dfrac{1}{2022}}\)
19 tháng 4 2022 lúc 19:04
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{\dfrac{2020}{1}+\dfrac{2019}{2}+\dfrac{2018}{3}+...+\dfrac{1}{2021}}\)
chi tiết nghen:))
9 tháng 5 2022 lúc 9:56
Tìm x, biết:
( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2023}\) ) . x = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) + \(\dfrac{2020}{3}\)
+ ... + \(\dfrac{1}{2022}\)
8 tháng 4 2022 lúc 11:03
tìm x:
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
Lưu ý: có cả cách giải:>
3 tháng 3 2023 lúc 22:57
So sánh:
a) A=\(\dfrac{98^{88}+1}{98^{98}+1}\)và B=\(\dfrac{98^{89}+1}{98^{99}+1}\) b) C=\(\dfrac{2022^{2023}+1}{2022^{2021}+1}\)và D=\(\dfrac{2022^{2021}+1}{2022^{2019}+1}\)
So sánh:
A = \(\dfrac{2^{2020}-1}{2^{2021}-1}\) và B = \(\dfrac{2^{2021}-1}{2^{2022}-1}\)
Câu 5 : A dfrac{1}{2} +dfrac{1}{2^2}+ dfrac{1}{2^3}+ dfrac{1}{2^4}+ ....+dfrac{1}{2^{2021}}+dfrac{1}{2^{2022}}và B dfrac{1}{3}+dfrac{1}{4}+dfrac{1}{5}+dfrac{17}{60}a) Rút gọn Ab) So sánh A và B
Đọc tiếp
Câu 5 : A= \(\dfrac{1}{2}\) +\(\dfrac{1}{2^2}\)+ \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)+ ....+\(\dfrac{1}{2^{2021}}\)+\(\dfrac{1}{2^{2022}}\)và B= \(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)+\(\dfrac{17}{60}\)
a) Rút gọn A
b) So sánh A và B
so sánh
a)A=\(\dfrac{17^{18}+1}{17^{19}+1}\)và B=\(\dfrac{17^{17}+1}{17^{18}+1}\)
b)C=\(\dfrac{2^{2020}-1}{2^{2021}-1}\)và D=\(\dfrac{2^{2021}-1}{2^{2022}-1}\)
c)\(\dfrac{13579}{34567}\)và \(\dfrac{13580}{34569}\)
Giúp mình với nhé😌