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A=1-1/2019+1-1/2020+1+2/2018
=>A=(1+1+1)+(1/2018-1/2009)+(1/2018-1/2020)
Vì 1/2018>1/2019 và 1/2028>1/2020
=>A>3
Vậy a >A
study well
k nha ủng hộ mk nhé
Mình cũng làm giống thế . nhưng con bạn mình làm a < 3 nên mình không chắc chắn
Ta có : \(\frac{1}{n}+\frac{2020}{2019}=\frac{2019}{2018}+\frac{1}{n+1}\)
=> \(\frac{1}{n}-\frac{1}{n+1}=\frac{2019}{2018}-\frac{2020}{2019}\)
=> \(\frac{n+1}{n\left(n+1\right)}-\frac{n}{\left(n+1\right)n}=\frac{1}{4074342}\)
=> \(\frac{1}{n\left(n+1\right)}=\frac{1}{2018.2019}\)
=> n(n + 1) = 2018.2019
=> n(n + 1) = 2018.(2018 + 1)
=> n = 2018
\(\left(2020\frac{2018}{2021}-2019\frac{20182018}{20212021}\right):\frac{2018}{2021}\)
\(=\left(2020\frac{2018}{2021}-2019\frac{2018}{2021}\right):\frac{2018}{2021}\)
\(=1:\frac{2018}{2021}=\frac{2021}{2018}\)
\(A=\frac{2020}{2019}-\frac{2019}{2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018}{2019\times2018}-\frac{2019\times2019}{2019\times2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018-2019\times2019+1}{2019\times2018}\)
\(=\frac{\left(2019+1\right)\times\left(2019-1\right)-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-2019+2019-1-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-1-\left(2019\times2019-1\right)}{2019\times2018}\)
\(=\frac{0}{2019\times2018}\)
\(=0\)
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
\(\frac{2019}{2020}+\frac{2020}{2019}=1-\frac{1}{2020}+1+\frac{1}{2019}\)
\(=2+\frac{1}{2019}-\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\Rightarrow\frac{1}{2019}-\frac{1}{2020}>0\)
\(\Rightarrow2+\frac{1}{2019}-\frac{1}{2020}>2\)
\(\frac{444443}{222222}=\frac{444444}{222222}-\frac{1}{222222}=2-\frac{1}{222222}< 2\)
\(\Rightarrow\frac{2019}{2020}+\frac{2020}{2019}>\frac{444443}{222222}\)
https://olm.vn/hoi-dap/detail/224964577156.html
THAM-KHẢO-NHÉ
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
ta có
\(1-\frac{2018}{2019}=\frac{1}{2019}\)và\(1-\frac{2019}{2020}=\frac{1}{2020}\)
vì\(\frac{1}{2019}>\frac{1}{2020}\)vậy\(\frac{2018}{2019}>\frac{2019}{2020}\)
a) Ta có \(\frac{13}{7}=2-\frac{1}{7}\)
\(\frac{21}{12}=2-\frac{1}{4}\)
Vì \(\frac{1}{7}< \frac{1}{4}\)\(\Rightarrow2-\frac{1}{7}>2-\frac{1}{4}\)\(\Rightarrow\frac{13}{7}>\frac{21}{12}\)
Vậy \(\frac{13}{7}>\frac{21}{12}\)
b) Ta có : \(\frac{2018}{2019}=1-\frac{1}{2019}\)
\(\frac{2019}{2020}=1-\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\Rightarrow1-\frac{1}{2019}< 1-\frac{1}{2020}\Rightarrow\frac{2018}{2019}< \frac{2019}{2020}\)
Vậy \(\frac{2018}{2019}< \frac{2019}{2020}\)
c) Ta có :Vì \(\frac{17}{53}< \frac{17}{50}< \frac{19}{50}\) \(\Rightarrow\frac{17}{53}< \frac{19}{50}\)
Vậy \(\frac{17}{53}< \frac{19}{50}\)
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
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