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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
\(A=\frac{2020}{2019}-\frac{2019}{2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018}{2019\times2018}-\frac{2019\times2019}{2019\times2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018-2019\times2019+1}{2019\times2018}\)
\(=\frac{\left(2019+1\right)\times\left(2019-1\right)-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-2019+2019-1-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-1-\left(2019\times2019-1\right)}{2019\times2018}\)
\(=\frac{0}{2019\times2018}\)
\(=0\)
Vậy A = 0
ta có
A=2020*2018/2019*2018-2019*2019/2018*2019+1/2018*2019
=>A*(2018*2019)=2020*2018-2019*2019+1
=>A*(2018*2019)=(2019+1)*2018-(2018+1)*2019+1
=>A*(2018*2019)=(2019*2018+2018)-(2018*2019+2019)+1
=>A*(2018*2019)=2019*2018+2018-2018*2019-2019+1
=>A*(2018*2019)=2018-2019+1
=>A*(2018*2019)=2018+1-2019
=>A*(2018*2019)=0
=>A=0/(2018*2019)
=>A=0
Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{138}{30}< X< \dfrac{200}{3}\)
\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)
Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
a: Số cần tìm là 5,32:0,125=42,56
b: \(A=1+\dfrac{1}{2019}-1-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}=0\)