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\(A=\frac{2020}{2019}-\frac{2019}{2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018}{2019\times2018}-\frac{2019\times2019}{2019\times2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018-2019\times2019+1}{2019\times2018}\)
\(=\frac{\left(2019+1\right)\times\left(2019-1\right)-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-2019+2019-1-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-1-\left(2019\times2019-1\right)}{2019\times2018}\)
\(=\frac{0}{2019\times2018}\)
\(=0\)
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
Ta có:
A = \(\dfrac{2017}{2019}=1-\dfrac{2}{2019}\)
B= \(\dfrac{2019}{2021}\) = 1- \(\dfrac{2}{2021}\)
Ta có:
\(\dfrac{2}{2019}>\dfrac{2}{2021}\)
=> 1- \(\dfrac{2}{2019}< 1-\dfrac{2}{2021}\)
=> \(\dfrac{2017}{2019}< \dfrac{2019}{2021}\)
Lại có \(\dfrac{1}{2}< \dfrac{2}{3}\)
=>\(\dfrac{2017}{2019}+\dfrac{1}{2}< \dfrac{2019}{2021}+\dfrac{2}{3}\)
Vậy A<B
\(A=\frac{2016^{2016}+1}{2016^{2017}+1}\Rightarrow2016A=\frac{2016^{2017}+2016}{2016^{2017}+1}=1+\frac{2015}{2016^{2017}+1}\)
\(B=\frac{2016^{2017}-3}{2016^{2018}-3}\Rightarrow2016B=\frac{2016^{2018}-6048}{2016^{2018}-3}=1+\frac{-6045}{2016^{2018}-3}\)
Vì \(\frac{2015}{2016^{2017}+1}>0;\frac{-6045}{2016^{2018}-3}< 0\)
Nên: A>B
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 :
\(\frac{2017\times2018+1}{2019+2016\times2018}\)
\(=\frac{2017\times2018+1}{1+2018+2016\times2018}\)
\(=\frac{2017\times2018+1}{1+2018\times\left(2016+1\right)}\)
\(=\frac{2017\times2018+1}{1+2018\times2017}\)
\(=1\)
\(\frac{2017.2018+1}{2019+2016.2018}\)
\(=\frac{2017.2018+1}{1+2018+2016.2018}\)
\(=\frac{2017.(2018+1)}{(1+2018).\left(2016+1\right)}\)
\(=\frac{2017.2019}{2019.2017}\)
\(=\frac{1}{1}=1\)
vi 2018/2019<1
2019/2020<1
2020/2021<1
nen 2018/2019 + 2019/2020 + 2020/2021<1+1+1=3
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+....+\frac{2019}{2018.2019}\)
\(=\frac{2019}{1}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\frac{2018}{2019}\)
\(=2018\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+\frac{2019}{20}+\frac{2019}{30}+\frac{2019}{2018.2019}\)
\(A=\frac{2019}{1.2}+\frac{2019}{2.3}+\frac{2019}{3.4}+\frac{2019}{4.5}+\frac{2019}{5.6}+...+\frac{2019}{2018.2019}\)
\(A=2019.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(A=2019.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=2019.\left(1-\frac{1}{2019}\right)\)\(=2019.\frac{2018}{2019}=2018\)
Vậy A = 2018
-Dấu " . " là dấu nhân.