Ta có:2019>4
=>2019/2020+2020/2021+2021/2022+2019>4
=>a>4(dpcm)
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 \(\dfrac{B}{A}\)
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:>>
tìm x:
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
Lưu ý: có cả cách giải:>
So sánh A=\(\dfrac{2018}{2019}\)+\(\dfrac{2019}{2020}\)+\(\dfrac{2020}{2021}\)+\(\dfrac{2021}{2018}\)với 4
a) Tìm số tự nhiên n biết:
\(\dfrac{4}{3\cdot5}+\dfrac{8}{5\cdot9}+\dfrac{12}{9\cdot15}+....+\dfrac{32}{n\cdot\left(n+16\right)}=\dfrac{16}{25}\)
b) Chứng tỏ rằng:
\(\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2018}>4\)
A = \(\dfrac{2022}{2021^{2^{ }}+1}\) + \(\dfrac{2022}{2021^{2^{ }}+2}\) + \(\dfrac{2022}{2021^2+3}\) + ... + \(\dfrac{2022}{2021^{2^{ }}+2021}\)
Chứng tỏ rằng A không phải số tự nhiên
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 P=2019/2020+2020/2021+2021/2022 và Q=2019+2020+2021/2020+2021+2022
So sánh A và B
A=\(\dfrac{4-7^{2020}}{7^{2020}}\)+\(\dfrac{5+7^{2021}}{7^{2021}}\)
B=\(\dfrac{1}{7^{2019}}\)
