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a: Số cần tìm là 5,32:0,125=42,56
b: \(A=1+\dfrac{1}{2019}-1-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}=0\)
\(A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)
Ta có: \(1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)\)
\(=\left(1+1+1+...+1\right)+\left(2+2+2+...+2\right)+\left(3+3+3+...+3\right)+...+\left(2019+2019\right)+2020\)
Trong đó có: 2020 số 1, 2019 số 2, 2018 số 3,..., 2 số 2019, 1 số 2020
Vậy: \(\left(1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+3+...+3\right)+...+2020\)
\(=1\times2020+2\times2019+3\times2018+...+2020\times1\)
\(\Rightarrow A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)
\(A=\dfrac{1\times2020+2\times2019+3\times2018+...+2020\times1}{1\times2020+2\times2019+3\times2018+...+2020\times1}=1\)
Ta có :
\(A=\dfrac{2019\times2020}{2019\times2020+1}=\dfrac{2019\times2020+1-1}{2019\times2020+1}=1-\dfrac{1}{2019\times2020+1}\)
Suy ra A < 1 (1)
Lại có \(B=\dfrac{2020}{2019}=\dfrac{2019+1}{2019}=\dfrac{2019}{2019}+\dfrac{1}{2019}=1+\dfrac{1}{2019}\)
Suy ra B > 1 (2)
Từ (1) và (2) ta có : A < 1 < B
=> A < B
Vậy A < B
2020/2019 x 2019/2018 x 2018/2017 x....................3/2
= 2020/2
= 1010
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
ét o ét