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b,
\(B=\frac{2018+2019}{2019+2020}=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)
Ta thấy :
\(\frac{2018}{2019}>\frac{2018}{2019+2020}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020}\)
Từ đó , suy ra :
\(\frac{2018}{2019}+\frac{2019}{2020}>\frac{2018+2019}{2019+2020}\)
Vậy...
#Louis
Ta có :
\(\frac{205}{321}< \frac{205}{315}\)
\(\frac{214}{315}>\frac{205}{315}\)
\(\Leftrightarrow\frac{205}{321}< \frac{214}{315}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có A= 27.58-31
=> A= (26+1).58-31
=> A= 26.58+58-31
=> A= 26.58+27 = B
Vậy A=B
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A =1987657 . 1987655
A =(1987656+1).1987655
A =1987656.1987655+1987655
B =1987656 . 1987656
B =(1987655+1).1987656
B =1987655.1987656+1987656
Vì 1987656.1987655=1987655.1987656 và 1987655<1987656
nên A<B
Ta có: 2.2 >1.3 ; 4.4 > 3.5
=> 1987657 . 1987655 < 1987656 . 1987656
Vậy A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
B=(454-1)(137+1)=137.454+454-137-1=137.454+316>137.454+206=A
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=2021\cdot\left(2021-1\right)\left(2021+1\right)=2021^3-1< 2021^3=A\)
Ta có:
20212 = 2021 . 2021 = 2021 + 2021 . 2020
2020 . 2022 = 2020 + 2020 . 2021
Nhận xét
2021 > 2020
2021 . 2020 = 2020 . 2021
=> 20212 > 2020 . 2022
=> 20213 > 2020 . 2021 . 2022
Hay A > B
dấu >
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