Cho f(x)=x^100-2020x^99+2020x^98-...+2020x^2-2020x+2000
Tính f(2019)
K
Khách
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NT
0
TV
2
26 tháng 6 2020
Bài làm:
Ta có: \(x=2019\Rightarrow2020=x+1\)
Thay vào ta được:
\(f\left(2019\right)=x^{99}-\left(x+1\right)x^{98}+\left(x+1\right)x^{97}-\left(x+1\right)x^{96}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(f\left(2019\right)=x^{99}-x^{99}-x^{98}+x^{98}+x^{97}-x^{97}-x^{96}+...-x^3-x^2+x^2+x-1\)
\(f\left(2019\right)=x-1\)
Thay \(x=2019\)vào ta được:
\(f\left(2019\right)=2019-1=2018\)
Vậy f(2019) = 2018
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
26 tháng 6 2020
\(f\left(x\right)=x^{99}-2020x^{98}+2020x^{97}-2020x^{96}+...-2020x^2+2020x-1\)
\(f\left(2019\right)=2019^{99}-2020.2019^{98}+2020.2019^{97}-...+2020.2019-1\)
Xét \(2020.2019^{98}=2019^{99}+2019^{98};2020.2019^{97}=2019^{98}+2019^{97}\)
\(2020.2019^{96}=2019^{97}+2019^{96};...;2020.2019=2019^2+2019\)
\(\Rightarrow f\left(2019\right)=2019^{99}-2019^{99}-2019^{98}+2019^{97}-2019^{97}-...+2019^2+2019-1\)
\(\Rightarrow f\left(2019\right)=2019-1=2018\). Vậy \(f\left(2019\right)=2018\)
VN
1
21 tháng 6 2021
f(x) = \(\left(x^6-2019x^5\right)-\left(x^5-2019x^4\right)+\left(x^4-2019x^3\right)-\left(x^3-2019x^2\right)+\left(x^2-2019x\right)-\left(x-2019\right)+1\)
= \(x^5\left(x-2019\right)-x^4\left(x-2019\right)+x^3\left(x-2019\right)-x^2\left(x-2019\right)+x\left(x-2019\right)-\left(x-2019\right)+1\)
Thay x = 2019 vào f(x), ta có:
f(2019) = 0 + 0 + 0 + 0 + 0 +0 + 1 = 1
PV
0
16 tháng 9 2021
\(D=4x^2-2x+3x\left(x-5\right)=4x^2-2x+3x^2-15x=7x^2-17x=7\left(-1\right)^2-17\left(-1\right)=24\)
\(E=x^{10}-2020x^9+2020x^8-2020x^7+...+2020x^2-2020x=x^9\left(x-2019\right)-x^8\left(x-2019\right)+x^7\left(x-2019\right)-...-x^2\left(x-2019\right)+x\left(x-2019\right)-x=x^9\left(2019-2019\right)-...+x\left(2019-2019\right)-2019=-2019\)
NP
0
NP
1
KN
8 tháng 11 2019
\(x^{2019}-2020x^{2018}+2020x^{2017}-2020x^{2016}+...+2020x-2020\)
\(=x^{2019}-2019x^{2018}-x^{2018}+2019x^{2017}+x^{2017}\)
\(-2019x^{2016}-x^{2016}+...+2019x+x-2020\)
\(=x^{2018}\left(x-2019\right)-x^{2017}\left(x-2019\right)+x^{2016}\left(x-2019\right)\)
\(+...-x\left(x-2019\right)+\left(x-2019\right)-1\)
\(=-1\)
VH
2
31 tháng 8 2018
2020.2019^5 = (2019+1).2019^5 = 2019^6+2019^5 làm tương tự với các x còn lại
A= 2019^6 - 2019^6 +.....-2019^2-2019 +2020 = 1 vậy A=1
\(f\left(2019\right)=x^{100}-\left(2019+1\right)x^{99}+\left(2019+1\right)x^{98}-....+\left(2019+1\right)x^2-\left(2019+1\right)x+2000\)
\(=x^{100}-\left(x+1\right)x^{99}+\left(x+1\right)x^{98}-...+\left(x+1\right)x^2-\left(x+1\right)x+2000\)
\(=x^{100}-x^{100}-x^{99}+x^{99}+x^{98}-...+x^3+x^2-x^2-x+2000\)
\(=-x+2000=-2019+2000\)
\(=-19\)