so sánh phân số 2023/2022 và 2021/2020
giúp mik vs đúng thì mik sẽ vote cho nhé =))
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Ta có:
\(A=\frac{2021^{2021}+1}{2021^{2022}+1}\Leftrightarrow10A=\frac{2021^{2022}+10}{2021^{2022}+1}=1+\frac{9}{2021^{2022}+1}\)
\(B=\frac{2021^{2022}-1}{2021^{2023}-1}\Leftrightarrow10B=\frac{2021^{2023}-10}{2021^{2023}-1}=1-\frac{9}{2021^{2023}-1}\)
Hay ta đang so sánh: \(\frac{9}{2021^{2022}};\frac{9}{2021^{2023}}\)
Mà \(\frac{9}{2021^{2022}}>\frac{9}{2021^{2023}}\)nên \(\frac{2021^{2021}+1}{2021^{2022}+1}>\frac{2021^{2022}-1}{2021^{2023}-1}\)hay\(A>B\)
Vậy \(A>B\)
Không cần tính, ta thấy : 2022/2021 > 2021/2022
Vậy : 2022/2021*2023 > 2021/2022*2022
a: \(B=\dfrac{154}{155+156}+\dfrac{155}{155+156}\)
\(\dfrac{154}{155}>\dfrac{154}{155+156}\)
\(\dfrac{155}{156}>\dfrac{155}{155+156}\)
=>154/155+155/156>(154+155)/(155+156)
=>A>B
b: \(C=\dfrac{2021+2022+2023}{2022+2023+2024}=\dfrac{2021}{6069}+\dfrac{2022}{6069}+\dfrac{2023}{6069}\)
2021/2022>2021/6069
2022/2023>2022/2069
2023/2024>2023/6069
=>D>C
1-2+3-4+...+2021-2022+2023
=(1-2)+(3-4)+...+(2021-2022)+2023
=(-1)+(-1)+(-1)+...+(-1)+2023
=(-1011)+2023
=1012
\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
mà 10^2023+1>10^2022+1
nên A<B
(125.72-5.352):20212022
=(53.72-53.72):20212022
=0:20212022
=0
100% ĐÚNG
\(\dfrac{2023}{2022}=\dfrac{2022}{2022}+\dfrac{1}{2022}=1+\dfrac{1}{2022}\)
\(\dfrac{2021}{2020}=\dfrac{2020}{2020}+\dfrac{1}{2020}=1+\dfrac{1}{2020}\)
\(\dfrac{1}{2022}< \dfrac{1}{2020}\)
\(\Rightarrow\dfrac{2023}{2022}< \dfrac{2021}{2020}\)
\(\dfrac{2023}{2022}=1+\dfrac{1}{2022}\)
\(\dfrac{2021}{2020}=1+\dfrac{1}{2020}\)
mà \(\dfrac{1}{2022}< \dfrac{1}{2020}\)
nên \(\dfrac{2023}{2022}< \dfrac{2021}{2020}\)