2020*2018+2019
__________________
2019*2019+2018
Tính nhanh
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\(\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2019}x2018\)
\(=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{2018}{2019}=2-\dfrac{2019}{2018}=\dfrac{2017}{2018}\)
\(\frac{2016}{2017}\)x \(\frac{2017}{2018}\)x \(\frac{2019}{2020}\)=\(\frac{504}{505}\)
đ/s:\(\frac{504}{505}\)
= (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
\(A=\dfrac{2020^{2018}-1}{2020^{2019}+2019}\)
\(B=\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
Ta có :
\(A-B=\dfrac{2020^{2018}-1}{2020^{2019}+2019}-\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
\(\Rightarrow A-B=\dfrac{\left(2020^{2018}-1\right)\left(2020^{2020}+2019\right)-\left(2020^{2019}+2019\right)\left(2020^{2019}+1\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{2020^{4038}+2019.2020^{2018}-2020^{2020}-2019-2020^{4038}-2020^{2019}-2019.2020^{2018}-2029}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{-\left(2020^{2020}+2020^{2019}+2.2019\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
mà \(\left\{{}\begin{matrix}-\left(2020^{2020}+2020^{2019}+2.2019\right)< 0\\\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)>0\end{matrix}\right.\)
\(\Rightarrow A-B< 0\)
\(\Rightarrow A< B\)
Vậy ta được \(A< B\)
Trả lời :...............................................
\(\frac{4078379}{4078379}\)
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