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NN
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SN
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BP
16 tháng 7 2023 lúc 21:28

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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NH
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HN
6 tháng 2 2023 lúc 1:11

\(T=\dfrac{1}{2^1}+\dfrac{2}{2^2}+...+\dfrac{2021}{2^{2021}}+\dfrac{2022}{2^{2022}}\)

\(\Leftrightarrow2T=1+\dfrac{1}{2}+\dfrac{3}{2^2}...+\dfrac{2020}{2^{2019}}+\dfrac{2021}{2^{2020}}+\dfrac{2022}{2^{2021}}\)

\(\Leftrightarrow2T-T=\left(1+\dfrac{1}{2}+\dfrac{3}{2^2}...+\dfrac{2020}{2^{2019}}+\dfrac{2021}{2^{2020}}+\dfrac{2022}{2^{2021}}\right)-\left(\dfrac{1}{2^1}+\dfrac{2}{2^2}+...+\dfrac{2021}{2^{2021}}+\dfrac{2022}{2^{2022}}\right)\)

\(\Leftrightarrow T=1+\dfrac{1}{2}+\dfrac{3}{2^2}...+\dfrac{2020}{2^{2019}}+\dfrac{2021}{2^{2020}}+\dfrac{2022}{2^{2021}}-\dfrac{1}{2^1}-\dfrac{2}{2^2}-...-\dfrac{2021}{2^{2021}}-\dfrac{2022}{2^{2022}}\)

\(\Leftrightarrow T=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2021}}-\dfrac{2022}{2^{2022}}\)

Đặt \(M=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2021}}\)

\(\Leftrightarrow2M=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\)

\(\Leftrightarrow2M-M=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2021}}\right)\)

\(\Leftrightarrow M=1-\dfrac{1}{2^{2021}}\)

Khi đó: \(T=1+M-\dfrac{2022}{2^{2022}}\)

\(\Leftrightarrow T=1+1-\dfrac{1}{2^{2021}}-\dfrac{2022}{2^{2022}}\)

\(\Leftrightarrow T=2-\left(\dfrac{1}{2^{2021}}+\dfrac{2022}{2^{2022}}\right)\)

\(Do\left(\dfrac{1}{2^{2021}}+\dfrac{2022}{2^{2022}}\right)>0\) \(nên\) \(suy\) \(ra\) \(T=2-\left(\dfrac{1}{2^{2021}}+\dfrac{2022}{2^{2022}}\right)< 2\)

Vậy \(T< 2\)           (\(ĐPCM\))

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MC
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LL
24 tháng 8 2021 lúc 17:06

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

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LH
24 tháng 8 2021 lúc 17:07

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

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TC
24 tháng 8 2021 lúc 17:07

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HL
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NT
25 tháng 6 2023 lúc 10:53

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PH
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MT
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NT
13 tháng 7 2023 lúc 11:34

\(B=\left(\dfrac{2020}{2}+1\right)+\left(\dfrac{2019}{3}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1\)

\(=\dfrac{2022}{2}+\dfrac{2022}{3}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}\)

=2022(1/2+1/3+...+1/2021+1/2022)

=>B/A=2022

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MN
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HN
16 tháng 4 2023 lúc 10:02

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

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NL
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NL
22 tháng 11 2021 lúc 20:50

\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+n^2+\left(n+1\right)^2}{n^2\left(n+1\right)^2}}\)

\(=\sqrt{\dfrac{\left(n^2+n\right)^2+n^2+n^2+2n+1}{\left(n^2+n\right)^2}}=\sqrt{\dfrac{\left(n^2+n\right)^2+2\left(n^2+n\right)+1}{\left(n^2+n\right)^2}}\)

\(=\sqrt{\dfrac{\left(n^2+n+1\right)^2}{\left(n^2+n\right)^2}}=\dfrac{n^2+n+1}{n^2+n}=1+\dfrac{1}{n\left(n+1\right)}\)

\(\Rightarrow A=1+\dfrac{1}{2.3}+1+\dfrac{1}{3.4}+....+1+\dfrac{1}{2021.2022}\)

\(=2020+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)

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

\(=2020+\dfrac{1}{2}-\dfrac{1}{2022}=...\)

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NM
22 tháng 11 2021 lúc 20:53

\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}=\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{3}}=\sqrt{\left(1+\dfrac{1}{2}-\dfrac{1}{3}\right)^2}=1+\dfrac{1}{2}-\dfrac{1}{3}\)

Cmttt ta được:

\(A=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}+1+\dfrac{1}{2021}-\dfrac{1}{2022}\\ A=2020+\dfrac{1}{2}-\dfrac{1}{2022}=2020+\dfrac{505}{1011}=...\)

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