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A = 2019 \(\times\) 2021 + 2023
A = (2018 + 1).(2022 -1) + 2023
A = 2018.2022 - 2018 + 2023 > 2018.2022 - 2022
Vậy A > B
Cách 1: Nhìn qua là biết A > B :))
Cách 2: Giải cụ thể:
A = 2019 x 2021 + 2023
= 2018 x 2021 + 2021 + 2023 = 2018 x 2021 + 4044
B = 2018 x 2022 - 2022
= 2018 x 2021 + 2018 - 2022 = 2018 x 2021 - 4
⇒ A > B và lớn hơn: 4044 + 4 = 4048
Ta có
A = 2017/2019 =1 - 2/2019
B = 2021/2023 = 1 - 2/2013
MÀ 2/2019 < 2/2013 => 1 - 2/2019 > 1 - 2/2013 hay A > B
Vậy A > B
Easy mà bạn :
Ta có :
\(A=\frac{2017}{2019}=1-\frac{2}{2019}\)
\(B=\frac{2021}{2023}=1-\frac{2}{2023}\)
Do \(\frac{2}{2019}>\frac{2}{2023}\)
\(\Rightarrow1-\frac{2}{2019}< 1-\frac{2}{2023}\)
\(\Rightarrow A< B\)
~
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
\(A=\dfrac{2019\times2021-1}{2019\times2021}=\dfrac{2019\times2021}{2019\times2021}-\dfrac{1}{2019\times2021}=1-\dfrac{1}{2019\times2021}\)
\(B=\dfrac{2021\times2023-1}{2021\times2023}=\dfrac{2021\times2023}{2021\times2023}-\dfrac{1}{2021\times2023}=1-\dfrac{1}{2021\times2023}\)
\(a)\dfrac{7}{8}=\dfrac{7\times9}{8\times9}=\dfrac{63}{72}\)
\(\dfrac{3}{9}=\dfrac{3\times8}{9\times8}=\dfrac{24}{72}\)
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 \(\dfrac{6}{7}>\dfrac{5}{6}\)
\(\dfrac{2021}{2019}và\dfrac{2023}{2021}\)
\(\Rightarrow\dfrac{2021}{2019}-\dfrac{2}{2019}=\dfrac{2023}{2021}-\dfrac{2}{2021}\left(=1\right)\)
\(\Rightarrow\dfrac{2}{2019}>\dfrac{2}{2021}\Rightarrow\dfrac{2021}{2019}< \dfrac{2023}{2021}\)
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 :
\(2021>2019\Rightarrow\dfrac{2021}{2019}>\dfrac{2021+2}{2019+2}=\dfrac{2023}{2021}\)