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Nhận xét : ( x + y - 3 )^2018 >=0 và 2018.(2x-4)^2020 >= 0
=> (x+y-3)^2018 + 2018.(2x-4)^2020 >=0
Dấu = xảy ra khi : x + y - 3 = 0 và 2x - 4 = 0 => x = 2 và y = 1
Thay vào bt S :
S = ( 2 - 1)^2019 + (2-1)^2019
= 1^2019 + 1^2019 = 2
\(A=1+3+3^2+3^3+3^4+3^5+.....+3^{2017}\)
\(=1+3+\left(3^2+3^3+3^4+3^5\right)+.....+\left(3^{2014}+3^{2015}+3^{2016}+3^{2017}\right)\)
\(=4+3^2\left(1+3+3^2+3^3\right)+.....+3^{2014}\left(1+3+3^2+3^3\right)\)
\(=4+3^2\cdot40+....+3^{2014}\cdot40\)
\(=4+40\left(3^2+.....+3^{2014}\right)\) chia 40 dư 4.
\(\frac{3-x}{2016}-1=\frac{2-x}{2017}+\frac{1-x}{2018}\)
\(\Rightarrow\frac{3-x}{2016}-1+2=\frac{2-x}{2017}+\frac{1-x}{2018}+2\)(thêm 2 vô mỗi vế)
\(\Rightarrow\frac{3-x}{2016}+1=\left(\frac{2-x}{2017}+1\right)+\left(\frac{1-x}{2018}+1\right)\)
\(\Rightarrow\frac{2019-x}{2016}=\frac{2019-x}{2017}+\frac{2019-x}{2018}\)
\(\Rightarrow\left(2019-x\right)\cdot\frac{1}{2016}=\left(2019-x\right)\left(\frac{1}{2017}+\frac{1}{2018}\right)\)
\(\Rightarrow2019-x=0\)
\(\Rightarrow x=2019\)
Gọi số người cần tìm là : a ( a < 1000 )
Theo đề bài, ta có :
(a - 15) chia hết cho 20;25;30
=> (a - 15) thuộc BC(20,25,30)
20 = 2^2 . 5
25 = 5^2
30 = 2.3.5
BCLN(20,25,30) = 2^2 .3.5 = 60
BC(20,25,30) = B(60) =(0,60,120,180,240,....,540,600)
=> a - 15 = (0,60,120,180,240,....,540,600,...)
a = (75,135,195,255,...,555,615,...)
vì a chia hết cho 41
=> a =615
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=-4\)
=> \(\left[\frac{x+1}{2020}+1\right]+\left[\frac{x+2}{2019}+1\right]+\left[\frac{x+3}{2018}+1\right]+\left[\frac{x+4}{2017}+1\right]=-4\)
=> \(\left[\frac{x+1}{2020}+\frac{2020}{2020}\right]+\left[\frac{x+2}{2019}+\frac{2019}{2019}\right]+\left[\frac{x+3}{2018}+\frac{2018}{2018}\right]+\left[\frac{x+4}{2017}+\frac{2017}{2017}\right]=-4\)
=> \(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=-4\)
=> \(\left[x+2021\right]\left[\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\right]=-4\)
Do \(\frac{1}{2020}>\frac{1}{2019}>\frac{1}{2018}>\frac{1}{2017}\)nên \(\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\ne0\)
Do đó : x + 2021 = -4 => x = -4 - 2021 = -2025
