2021 + 2022 + 2019 . 2 : ( 2 - 2 )
Những câu hỏi liên quan
1. So sánh
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\) và B= \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{13}{60}\)
b) \(C=\dfrac{2019}{2021}+\dfrac{2021}{2022}\) và \(D=\dfrac{2020+2022}{2019+2021}.\dfrac{3}{2}\)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
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Câu 24: Cho biểu thức: A=1/2+1/3+1/4+.........+1/2021+1/2022 Và B=2021/1+2020/2+2019/3+.........+3/2019+2020+1/2021
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
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27 tháng 3 2022 lúc 21:50
Cho A = 1/2+1/3+1/4+...+1/2022 và
B = 2021/1+2020/2+2019/3+...+1/2021.
hứa cho đúng
Bạn kiểm tra đề giúp mình! Bạn yêu cầu gì về giả thiết trên?
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so sánh P=2019/2020+2020/2021+2021/2022 và Q=2019+2020+2021/2020+2021+2022
1+2-3-4+5-6-7-8+...-2019-2020+2021+2022
=1+(2-3-4+5)+(6-7-8+9)+.....+(2018-2019-2020+2021)+2022
=1+0+0+.....+0+2022
=2023
số năm nay luôn
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1+2-3-4+5+6-7-8-....-2019-2020+2021+2022 help
Ta có: 1+2-3-4+5+6-7-8+.....-2019-2020+2021+2022
=1+(2-3-4+5)+(6-7-8+9)+.....+(2018-2019-2020+2021)+2022
=1+0+0+.....+0+2022
=2023
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x-4/2022+x-3/2021+x-2/2020+x-1/2019=-4
\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}\text{=}-4\)
\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}+4\text{=}0\)
\(\left(\dfrac{x-4}{2022}+1\right)+\left(\dfrac{x-3}{2021}+1\right)+\left(\dfrac{x-2}{2020}+1\right)+\left(\dfrac{x-1}{2019}+1\right)\text{=}0\)
\(\dfrac{x-2018}{2022}+\dfrac{x-2018}{2021}+\dfrac{x-2018}{2020}+\dfrac{x-2018}{2019}\text{=}0\)
\(\left(x-2018\right)\left(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\right)\text{=}0\)
\(Do:\) \(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\ne0\)
\(x-2018\text{=}0\)
\(x\text{=}2018\)
\(Vậy...\)
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tính tổng sau : 7^2022-7^2021+7^2020-7^2019+...+7^2-7
\(A=7^{2022}-7^{2021}+7^{2020}-7^{2019}+...+7^2-7\)
\(\Rightarrow7A=7^{2023}-7^{2022}+7^{2021}-...+7^3-7^2\)
\(\Rightarrow8A=A+7A=7^{2022}-7^{2021}+...+7^2-7+7^{2023}-7^{2022}+...+7^3-7^2=7^{2023}-7\)
\(\Rightarrow A=\dfrac{7^{2023}-7}{8}\)
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1-2-3+4+5-6-7+8+...+2018-2019-2020+2021+2022-2023
Sửa đề: 1-2-3+4+5-6-7+8+...-2018-2019+2020+2021-2022-2023
=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+(2021-2022-2023)
=0+0+...+0+(-1-2023)
=-2024
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