so sánh :
4/9 và 13/18 , 34/-4 và -8,6 , 2021/2022 và 2022/2023
giải thích
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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
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022
B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\)
B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\)
B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))
Vậy B > C
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) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
Ta có:
\(2023^{2022}=2023\cdot2023^{2021}\)
\(2022^{2022}+2022^{2021}=2022^{2021}\cdot\left(2022+1\right)=2023\cdot2022^{2021}\)
Mà: \(2023>2022\)
\(\Rightarrow2023^{2021}>2022^{2021}\)
\(\Rightarrow2023^{2021}\cdot2023>2022^{2021}\cdot2023\)
\(\Rightarrow2023^{2022}>2022^{2022}+2022^{2021}\)
Vậy: ...
\(a)\dfrac{7}{8}=\dfrac{7\times9}{8\times9}=\dfrac{63}{72}\)
\(\dfrac{3}{9}=\dfrac{3\times8}{9\times8}=\dfrac{24}{72}\)
Do : \(\dfrac{63}{72}>\dfrac{24}{72}\) nên \(\dfrac{7}{8}>\dfrac{3}{9}\)
Không thì bạn có thể rút gọn 3/9 đi làm cho nó gọn ạ.
\(b)\) Ta thấy : \(\dfrac{2023}{2021}>1\) ( vì tử lớn hơn mẫu )
\(\dfrac{2021}{2022}< 1\) ( vì tử bé hơn mẫu )
Do đó : \(\dfrac{2023}{2021}>\dfrac{2021}{2022}\)
\(c)\dfrac{5}{6}=\dfrac{5\times7}{6\times7}=\dfrac{35}{42}\)
\(\dfrac{6}{7}=\dfrac{6\times6}{7\times6}=\dfrac{36}{42}\)
Do : \(\dfrac{36}{42}>\dfrac{35}{42}\) nên \(\dfrac{6}{7}>\dfrac{5}{6}\)
2023/2022=1+1/2022
2022/2021=1+1/2021
mà 2022>2021
nên 2023/2022<2022/2021
a: \(\dfrac{4}{9}=\dfrac{4\cdot2}{9\cdot2}=\dfrac{8}{18}< \dfrac{13}{18}\)
b: 34/-4=-8,5
Ta có: 8,5<8,6
=>-8,5>-8,6
=>\(\dfrac{34}{-4}>-8,6\)
c: \(\dfrac{2021}{2022}=1-\dfrac{1}{2022}\)
\(\dfrac{2022}{2023}=1-\dfrac{1}{2023}\)
Ta có: 2022<2023
=>\(\dfrac{1}{2022}>\dfrac{1}{2023}\)
=>\(-\dfrac{1}{2022}< -\dfrac{1}{2023}\)
=>\(-\dfrac{1}{2022}+1< -\dfrac{1}{2023}+1\)
=>\(\dfrac{2021}{2022}< \dfrac{2022}{2023}\)
34/-4=-8,5 là sao v