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\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\left(\dfrac{2020-1}{2019}\right)-\left(\dfrac{2019-1}{2018}\right)\)
\(A=1-1\)
\(A=0.\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\dfrac{2019}{2019}-\dfrac{2018}{2018}\)
\(A=1-1\)
\(A=0\)
= (1-1/2018)-(1+1/2018)-2020/2019
= 1-1/2018-1-1/2018-2020/2019
= -2/2018-2020/2019
vậy thôi
=(1-1/2018)-(1+1/2018)-2020/2019
=1-1/2018-1-1/2018-2020/2019
=-2/2018-2020/2019
Trả lời :...............................................
\(\frac{4078379}{4078379}\)
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé Kim Râu La
\(\frac{2016}{2017}\)x \(\frac{2017}{2018}\)x \(\frac{2019}{2020}\)=\(\frac{504}{505}\)
đ/s:\(\frac{504}{505}\)
Trả lời:
\(A=\frac{2}{2018.2020}+\frac{2021}{2020}-\frac{2020}{2019}\)
\(A=\frac{1}{2018}-\frac{1}{2020}+1+\frac{1}{2020}-\left(1+\frac{1}{2018}\right)\)
\(A=\frac{1}{2018}-\frac{1}{2020}+1+\frac{1}{2020}-1-\frac{1}{2018}\)
\(A=0\)
\(A=\frac{2}{2018}\cdot2020+\frac{2021}{2020}-\frac{2019}{2018}\)
\(A=\frac{2\cdot2020-2019}{2018}+\frac{2021}{2020}\)
\(A=\frac{2021}{2018}+\frac{2021}{2020}\)
\(A=\frac{2021\cdot\left(2020+2018\right)}{2018\cdot2020}=\frac{2021\cdot4038}{2018\cdot2020}=\frac{2021\cdot2019\cdot2}{2018\cdot1010\cdot2}=\frac{2020^2-1}{2018\cdot101\cdot10}\)
\(A=\frac{4080399}{20200180}\)
tớ nghĩ là các phân số trên đều là những ps nhỏ hơn 1 lên A<3
mình chỉ nghĩ thôi. K biết đúng hay sai đâu. đúng thì tích còn sai thì bỏ qua cho
https://olm.vn/hoi-dap/detail/224964577156.html
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THANKS
Ta có: \(\frac{2018}{2019}\)+ \(\frac{2019}{2020}\)+\(\frac{2020}{2018}\)= (1-\(\frac{1}{2019}\)) + ( 1 -\(\frac{1}{2020}\)) + ( 1 - \(\frac{1}{2018}\)) = ( 1+1+1) - (\(\frac{1}{2019}+\frac{1}{2020}+\frac{1}{2018}\)) = 3 - (\(\frac{1}{2019}+\frac{1}{2020}+\frac{1}{2018}\)) \(\Leftrightarrow\)3 - (\(\frac{1}{2019}+\frac{1}{2020}+\frac{1}{2018}\)) <3 Vậy \(\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2018}\)< 3
\(\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2019}x2018\)
\(=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{2018}{2019}=2-\dfrac{2019}{2018}=\dfrac{2017}{2018}\)