tinh p=x\(^{15}\)-2018x\(^{14}\)+2018x\(^{13}\)-2018x\(^{12}\)+...+2018x\(^3\)-2018x\(^2\)+2018x-2018 ;voi x=2017
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\(A=x^9-2018x^8+2018x^7-2018x^6+2016x^5-2018x^4+2018x^3-2018x^2+2018x-2018\)
\(A=x^9-\left(2017+1\right)x^8+\left(2017+1\right)x^7-...+\left(2017+1\right)x-\left(2017+1\right)\)
\(A=x^9-\left(x+1\right)x^8+\left(x+1\right)x^7-...+\left(x+1\right)x-x-1\)
\(A=x^9-x^9-x^8+x^8+x^7-...+x^2+x-x-1\)
\(A=-1\)
F(x)=\(x^7-2018x^6+2018x^5-2018x^4+2018x^3-2018x^2+2018x+1.\)
x=2017=>2018=x+1 thay vào F(x) ta có:
F(x)=x+1=2018
Lời giải:
Ta có:
\(A=x^5-2018x^4+2018x^3-2018x^2+2018x-1000\)
\(A=(x^5-2017x^4)-(x^4-2017x^3)+(x^3-2017x^2)-(x^2-2017x)+x-1000\)
\(A=x^4(x-2017)-x^3(x-2017)+x^2(x-2017)-x(x-2017)+x-1000\)
Tại \(x=2017\Rightarrow A=2017^4.0-2017^3.0+2017^2.0-2017.0+2017-1000\)
\(A=2017-1000=1017\)
Vì \(x=2017\Rightarrow x+1=2018\)
Thay \(x+1=2018\)vào biểu thức A ta được :
\(A=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-...-\left(x+1\right)x+\left(x+1\right)\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-...-x^2-x+x+1\)
\(=1\)
\(P\left(x\right)=x^{2017}-2018x^{2017}+2018x^{2016}-...-2018x+1\)
Vì \(x=2017\)
\(\Leftrightarrow x+1=2018\)
Thay vào P(x) ta được :
\(P\left(x\right)=x^{2017}-x^{2017}\left(x+1\right)+x^{2016}\left(x+1\right)-...-x\left(x+1\right)+1\)
\(P\left(x\right)=x^{2017}-x^{2018}-x^{2017}+x^{2017}+x^{2016}-...-x^2-x+1\)
\(P\left(x\right)=-x^{2018}+1\)
\(P\left(x\right)=-2017^{2018}+1\)
Gợi ý nhé bạn:
Xét x lớn hơn hoặc bằng 1
Ta có
2018x+2018x+y\(\ge\)2018+2018+y >225
=> vô lí
=> x=0
Xét x=0 vào biểu thức trên
Ta có: x=2017
nên x+1=2018
Ta có: \(P=x^{15}-2018x^{14}+2018x^{13}-2018x^{12}+...+2018x^3-2018x^2+2018x-2018\)
\(=x^{15}-\left(x+1\right)\cdot x^{14}+\left(x+1\right)\cdot x^{13}-\left(x+1\right)\cdot x^{12}+...+\left(x+1\right)\cdot x^3-\left(x+1\right)\cdot x^2+\left(x+1\right)\cdot x-\left(x+1\right)\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}+...+x^3-x^3+x^2-x^2+x-x-1\)
=-1
@ 肖战Daytoy_1005 giup