Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(f\left(2019\right)=x^{100}-\left(2019+1\right)x^{99}+\left(2019+1\right)x^{98}-....+\left(2019+1\right)x^2-\left(2019+1\right)x+2000\)
\(=x^{100}-\left(x+1\right)x^{99}+\left(x+1\right)x^{98}-...+\left(x+1\right)x^2-\left(x+1\right)x+2000\)
\(=x^{100}-x^{100}-x^{99}+x^{99}+x^{98}-...+x^3+x^2-x^2-x+2000\)
\(=-x+2000=-2019+2000\)
\(=-19\)
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
\(f\left(x\right)=x^{99}-2020x^{98}+2020x^{97}-2020x^{96}+...-2020x^2+2020x-1\)
\(f\left(2019\right)=2019^{99}-2020.2019^{98}+2020.2019^{97}-...+2020.2019-1\)
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 \(f\left(2019\right)=2018\)
b) Ta có : \(x=2019\) \(\Rightarrow x+1=2020\) Thay vào biểu thức ta được :
( Chỗ nào có 2020 thay thành x + 1 )
\(x^9-\left(x+1\right).x^8+\left(x+1\right).x^7-....-\left(x+1\right).x^2+\left(x+1\right).x\)
\(=x^9-x^9-x^8+x^8+x^7-...-x^3-x^2+x^2+x\)
\(=x\\ \)
\(=2019\)
Vậy : biểu thức trên bằng 2019 với x = 2019.
Dễ thấy \(VT\ge0\Rightarrow2020x\ge0\Leftrightarrow x\ge0\)
\(\Rightarrow pt\Leftrightarrow2019x+\frac{2019.2020}{2}=2020x\)
\(\Leftrightarrow x=2019.1010\)
\(Q\left(x\right)=x^{101}-2020x^{100}-2022x^{99}+2022x^{98}+x-2021\)
\(=x^{100}\left(x-2021\right)+x^{99}\left(x-2021\right)-x^{98}\left(x-2021\right)+x^{98}+x-2021\)
\(Q\left(2021\right)=0+0-0+2021^{98}+0=2021^{98}\)
f(x) = x^6 - 2020x^5 + 2020x^4 - 2020x^3 + 2020x^2 - 2020x + 2020
f(x) = x^6 - (2019+1)x^5 + (2019+1)x^4 - (2019+1)x^3 + (2019+1)x^2 - (2019+1)x + (2019+1)
f(x)= x^6 - 2019x^5 -x^5+2019x^4 +x^4-2019x^3-x^3+2019x^2 +x^2-2019x-x+2019+1
f(2019)= 2019^6-2019.2019^5 -2019^5+2019.2019^4 +2019^4-2019.2019^3 -2019^3+2019.2019^2 +2019^2-2019.2019 -2019+2019+1
f(2019)=2019^6-2019^6-2019^5+2019^5 +2019^4-2019^4 -2019^3+2019^3 +2019^2-2019^2-2019 +2019+1=1
Vậy f(2019)=1
@꧁ミ〖★ Äŋħ ✔𝕽ҽäӀ✔⁀★〗ミ♪ ᶦᵈᵒᶫ꧂
\(f\left(x\right)=x^6-2020x^5+2020x^4-2020x^3+2020x^2-2020x+2020.\)
\(f\left(x\right)=x^6-\left(2019+1\right)x^5-\left(2019+1\right)x^4-\left(2019+1\right)\)
\(f\left(x\right)=x^6-2019x^5-x^5+2019x^4+x^4-2019x^3-x^3+2019x^2-x^2-2019x-x+2019+1\)
\(f\left(x\right)=2019^6-2019.2019^5+2019^5+2019.2019^4-2019^4+2019.2019^3-2019^3\)\(+2019.2019^2+2019^2+2019+2019+1\)
\(f\left(2019\right)=1\)