so sánh phân số: 2023/2021 và 19/17
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2021/2023=1-2/2023
101/103=1-2/103
mà 2/2023<2/103
nên 2021/2023>101/103
\(\dfrac{2023}{2022}=\dfrac{2022}{2022}+\dfrac{1}{2022}=1+\dfrac{1}{2022}\)
\(\dfrac{2021}{2020}=\dfrac{2020}{2020}+\dfrac{1}{2020}=1+\dfrac{1}{2020}\)
\(\dfrac{1}{2022}< \dfrac{1}{2020}\)
\(\Rightarrow\dfrac{2023}{2022}< \dfrac{2021}{2020}\)
\(\dfrac{2023}{2022}=1+\dfrac{1}{2022}\)
\(\dfrac{2021}{2020}=1+\dfrac{1}{2020}\)
mà \(\dfrac{1}{2022}< \dfrac{1}{2020}\)
nên \(\dfrac{2023}{2022}< \dfrac{2021}{2020}\)
a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)
b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)
c) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
a.2021/2023 < 2017/2019
b.2005/2007 > 2009/2011
Giải thích : So sánh mẫu số, phân số có mẫu số bé hơn thì nó lớn hơn
\(\dfrac{2021}{2019}và\dfrac{2023}{2021}\)
\(\Rightarrow\dfrac{2021}{2019}-\dfrac{2}{2019}=\dfrac{2023}{2021}-\dfrac{2}{2021}\left(=1\right)\)
\(\Rightarrow\dfrac{2}{2019}>\dfrac{2}{2021}\Rightarrow\dfrac{2021}{2019}< \dfrac{2023}{2021}\)
Chứng minh bđt phụ nếu a>b \(\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(vớim\in N^{\circledast}\right)\Rightarrow a\left(b+m\right)>b\left(a+m\right)\Rightarrow ab+am>ab+bm\Rightarrow am>bm\Rightarrow a>b\) \(\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(1\right)\)
Áp dụng bđt (1) có :
\(2021>2019\Rightarrow\dfrac{2021}{2019}>\dfrac{2021+2}{2019+2}=\dfrac{2023}{2021}\)
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
\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
mà 10^2023+1>10^2022+1
nên A<B
\(\dfrac{2023}{2021}< \dfrac{19}{17}\)
\(\dfrac{2023}{2021}< \dfrac{19}{17}\)
Chúc e học tốt nha!☘