So sánh:
\(A=\frac{2019}{2020}+\frac{2020}{2021}\) và \(B=\frac{2019+2020}{2020+2021}\)
Những câu hỏi liên quan
so sánh
a)
A=\(\frac{10^{2020}+1}{10^{2021}+1};B=\frac{10^{2021}+1}{10^{2022}+1}\)
b)
\(A=\frac{2019}{2020}+\frac{2020}{2021}\)và \(B=\frac{2019+2020}{2020+2021}\)
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
a) A=10^2020+1/10^2021+1 < 10^2020+1+9/10^2022+1+9 =
10.(10^2021+1)/10.(10^2022+1) = 10^2021+1/10^2022+1 = B
Vậy A < B.
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So sánh M và N
M=\(\frac{2019}{2020}\)+\(\frac{2020}{2021}\)và N=\(\frac{2019+2020}{2020+2021}\)
N =2019+2020/2020+2021
=2019/2020+2021 + 2020/2020+2021
Ta có:
2019/2020>2019/2020+2021
2020/2021 > 2020/2020+2021
=>M>N
Bài tập: So Sánh
M= \(\frac{2018}{2019}\)+\(\frac{2019}{2020}\)+\(\frac{2020}{2021}\)
N=\(\frac{2018+2019+2020}{2019+2020+2021}\)
Ta có :
\(N=\frac{2018+2019+2020}{2019+2020+2021}\)
\(=\frac{2018}{2019+2020+2021}+\frac{2019}{2019+2020+2021}+\frac{2020}{2019+2020+2021}\)
Mà \(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Leftrightarrow M>N\)
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Trả lời:
Ta có:
\(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Rightarrow\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2021}>\frac{2018+2019+2020}{2019+2020+2021}\)
hay \(M>N\)
Vậy \(M>N\)
Ta có :
N = \(\frac{2018}{2019+2020+2021}+\frac{2019}{2019+2020+2021}+\frac{2020}{2019+2020+2021}\)
Mà \(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Rightarrow M>N\)
so sánh P=2019/2020+2020/2021+2021/2022 và Q=2019+2020+2021/2020+2021+2022
17 tháng 4 2021 lúc 17:53
So sánh M = \(\dfrac{2019}{2020}+\dfrac{2020}{2021}\) và N = \(\dfrac{2019+2020}{2020+2021}\)
Giải:
Ta có: N=2019+2020/2020+2021
=>N=2019/2020+2021 + 2020/2020+2021
Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021
=>M>N
Vậy ...
Chúc bạn học tốt!
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Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
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So sánh
a)\(\frac{-60}{12}\)và -0,8
b) \(\frac{2020}{2019}\)và \(\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}\)và \(\frac{10^{2019}+1}{10^{2020+1}}\)
a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)
\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)
Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)
b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)
\(\frac{2021}{2020}=1+\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)
\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)
Đến đây tự so sánh rồi nhé
2019 x 2020 - 1/ 2019 x 2020 và 2020 x 2021 - 1/ 2020 x 2021
so sánh phân số
So sánh A= \(\frac{79^{2018}+1}{79^{2019}+1};B=\frac{79^{2019}-2021}{79^{2020}-2011}\)
LÀm đỡ mk tí mk ko có nhiều tgian vi còn 5 đề nữa
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so sánh: A=2019^2019+1/2019^2020+1 và B=2019^2020+1/2019^2021+1
Vì 2019 + 2020 < 2019 + 2021 nên A < B