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A = \(\dfrac{2020}{2019}\) - \(\dfrac{2019}{2018}\) + \(\dfrac{1}{2019\times2018}\)
A = \(\dfrac{2020}{2019}\) - \(\dfrac{2019}{2018}\) + ( \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\))
A = \(\dfrac{2020}{2019}\) - \(\dfrac{2019}{2018}\) + \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\)
A = ( \(\dfrac{2020}{2019}\) - \(\dfrac{1}{2019}\)) - ( \(\dfrac{2019}{2018}\) - \(\dfrac{1}{2018}\))
A = \(\dfrac{2019}{2019}\) - \(\dfrac{2018}{2018}\)
A = 1 - 1
A = 0
Ta có : \(\dfrac{2018}{2019}-\dfrac{2017}{2018}=\dfrac{2018.2018-2019.2017}{2018.2019}\)
\(=\dfrac{2018.2018-\left(2018+1\right).\left(2018-1\right)}{2018.2019}\)
\(=\dfrac{2018.2018-2018.2018+2018-2018+1}{2018.2019}\)
\(=\dfrac{1}{2018.2019}=\dfrac{1}{4074342}\)
\(\dfrac{2018}{2019}-\dfrac{2017}{2018}\)=\(\dfrac{1}{4074342}\)
2020/2019 x 2019/2018 x 2018/2017 x....................3/2
= 2020/2
= 1010
\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)(1)
\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)(2)
Từ(1) và (2)
\(\Rightarrow B>A\)
\(A=\frac{2017\times2108}{2017\times2018+1}=1-\frac{1}{2017\times2018+1}\)
\(B=\frac{2018\times2019}{2018\times2019+1}=1-\frac{1}{2018\times2019+1}\)
Nhận thấy:\(2017\times2018< 2018\times2019\)
=> \(2017\times2018+1< 2018\times2019+1\)
=> \(\frac{1}{2017\times2018+1}>\frac{1}{2018\times2019+1}\)
=> \(A< B\)
a: Số cần tìm là 5,32:0,125=42,56
b: \(A=1+\dfrac{1}{2019}-1-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}=0\)
b.\(\dfrac{1}{2019.2018}\)
b nhé
nhiên 5a1 dúng ko