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\(\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2019}x2018\)
\(=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{2018}{2019}=2-\dfrac{2019}{2018}=\dfrac{2017}{2018}\)
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
Trả lời :...............................................
\(\frac{4078379}{4078379}\)
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé Kim Râu La
Ta có:
\(A=\frac{2018+2019}{2019+2020}=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)
mà 2019+2020 >2019>2020 \(\Rightarrow\frac{2018}{2019+2020}< \frac{2018}{2019};\frac{2019}{2019+2020}< \frac{2019}{2020}\)
\(\Rightarrow\frac{2018}{2019+2020}+\frac{2019}{2019+2020}< \frac{2018}{2019}+\frac{2019}{2020}\)hay \(A< B\)
= (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
\(\frac{2016}{2017}\)x \(\frac{2017}{2018}\)x \(\frac{2019}{2020}\)=\(\frac{504}{505}\)
đ/s:\(\frac{504}{505}\)
\(\frac{2016}{2017}\times\frac{2017}{2018}\times\frac{2018}{2019}\times\frac{2019}{2020}\)=
\(0,998109801980198\)
Đổi ra ta sẽ có !
\(\frac{504}{505}\)
Vậy là : ...................
Ta có:
\(1-\frac{2017}{2018}=\frac{1}{2018};1-\frac{2018}{2019}=\frac{1}{2019};1-\frac{2019}{2020}=\frac{1}{2020}\)
Vì \(\frac{1}{2018}>\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2017}{2018}< \frac{2018}{2019}< \frac{2019}{2020}\)
2017/2018 = (2018-1)/2018 = 1-1/2018
2018/2019 = (2019-1)/2019 = 1 - 1/2019
2019/2020 = (2020-1)/2020 = 1 - 1/2020
Có 1/2018 > 1/2019 > 1/2020 => 2017/2018 < 2018/2019 < 2019/2020
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\left(\dfrac{2020-1}{2019}\right)-\left(\dfrac{2019-1}{2018}\right)\)
\(A=1-1\)
\(A=0.\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\dfrac{2019}{2019}-\dfrac{2018}{2018}\)
\(A=1-1\)
\(A=0\)