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Ta có: 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ... + ( 1 + 2 + 3 +...+ 2020)
= ( 1 + 1 + 1 +... + 1 ) + (2 + 2 +...+ 2 ) + ( 3 + 3+...+ 3 ) + ...+ 2020
Có 2020 số 1 ; 2019 số 2 ; 2018 số 3 ;... ; 1 số 2020
= 2020 x 1 + 2019 x 2 + 2018 x 3 + ... + 2020x 1
=> \(M=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+...+2020\times1}\)
= \(\frac{1\times2020+2\times2019+...+2020\times1}{1\times2020+2\times2019+...+2020\times1}=1\)
\(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\\ =\dfrac{2020\times\left(2018+2022\right)}{2020\times4040}\\ =\dfrac{2020\times4040}{2020\times4040}\\ =1\)
\(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\)
\(=\dfrac{2020\times\left(2018+2022\right)}{2020\times4040}\)
\(=\dfrac{2018+2022}{4040}\)
\(=\dfrac{4040}{4040}\)
\(=1\)
B=2021 x 13 + 2009 + 2020 x 2007/2020 + 2020 x520x2020
B=2021 x 13 +2009+2020/1x2007/2020 +2020x520x2020
B=26273+2009+20201/1x2007/20201+2121808000
B=28282+2007+2121808000
B=2121838289
Trả lời:
\(A=\frac{2}{2018.2020}+\frac{2021}{2020}-\frac{2020}{2019}\)
\(A=\frac{1}{2018}-\frac{1}{2020}+1+\frac{1}{2020}-\left(1+\frac{1}{2018}\right)\)
\(A=\frac{1}{2018}-\frac{1}{2020}+1+\frac{1}{2020}-1-\frac{1}{2018}\)
\(A=0\)
\(A=\frac{2}{2018}\cdot2020+\frac{2021}{2020}-\frac{2019}{2018}\)
\(A=\frac{2\cdot2020-2019}{2018}+\frac{2021}{2020}\)
\(A=\frac{2021}{2018}+\frac{2021}{2020}\)
\(A=\frac{2021\cdot\left(2020+2018\right)}{2018\cdot2020}=\frac{2021\cdot4038}{2018\cdot2020}=\frac{2021\cdot2019\cdot2}{2018\cdot1010\cdot2}=\frac{2020^2-1}{2018\cdot101\cdot10}\)
\(A=\frac{4080399}{20200180}\)
\(\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2019}x2018\)
\(=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{2018}{2019}=2-\dfrac{2019}{2018}=\dfrac{2017}{2018}\)
\(x\times2020-x=2020\times2018+2020\)
\(\Rightarrow x\times\left(2020-1\right)=2020\times\left(2018+1\right)\)
\(\Rightarrow x\times2019=2020\times2019\)
\(\Rightarrow x=2020\)
x * 2020 - x = 2020 * 2018 + 2020
x * 2020 - x * 1 = 2020 * 2018 + 2020 * 1
x * ( 2020 - 1) = 2020 * (2018 + 1)
x * 2019 = 2020 * 2019
Vậy x = 2020