P(x)= x^2017 - 2018x^2016+ 2018x^2015+...-2018x^2 + 2018x-1

=> P(x)= x^2017 -2017x^2016-x^2016 + 2017x^2015 + x^2015+..-2017x^2-x^2 + 2017x+x-1

=> P(x)= x^2016(x-2017) -x^2015(x-2017)+...- x(x -2017)+ x-1

thay x=2017 vào p(x) ta được

p(2017)= 2016

\(P\left(x\right)=x^{2017}-2018x^{2017}+2018x^{2016}-...-2018x+1\)

Vì \(x=2017\)

\(\Leftrightarrow x+1=2018\)

Thay vào P(x) ta được :

\(P\left(x\right)=x^{2017}-x^{2017}\left(x+1\right)+x^{2016}\left(x+1\right)-...-x\left(x+1\right)+1\)

\(P\left(x\right)=x^{2017}-x^{2018}-x^{2017}+x^{2017}+x^{2016}-...-x^2-x+1\)

\(P\left(x\right)=-x^{2018}+1\)

\(P\left(x\right)=-2017^{2018}+1\)

a)\(A=x^5-2018x^4+2018x^3-2018x^2+2018x-2019\)

\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-2019\)

\(A=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2019\)

\(A=x-2019=2017-2019=-2\)

b)ta có:\(\left(x+1\right)^{20}+\left(y+2\right)^{30}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)

Thay vào \(\Rightarrow B=2\cdot\left(-1\right)^5+5\cdot\left(-2\right)^3+4\)

\(B=-2+\left(-40\right)+4=-38\)

thục hiền đc đó thục hiền ak nay vẫn hoc24 bình thường à

Ta có x=2017 => 2018 = x+1 ; 2019= x+2

thay vào ta có : \(A=x^5-\left(x+1\right).x^4+\left(x+1\right).x^3-\left(x+1\right).x^2+\left(x+1\right).x-\left(x+2\right)\) \(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^x+x-x-2\) \(=\left(x^5-x^5\right)+\left(-x^4+x^4^{ }\right)+\left(x^3-x^3\right)+\left(-x^2+x^2\right)+\left(x-x\right)-2\)=-2

ey học tốt nhá

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

Đề:............

<=> - (1 - 2018x) + 2019x.(1 - 2018x) = 0

<=> (1 - 2018x).[(-1) + 2019x] = 0

Xét 2 trường hợp, ta có:

TH₁: 1 - 2018x = 0          TH₂: -1 + 2019x = 0

<=> 2018x = 1                 <=> 2019x = 1

<=> x = 1/2018                <=> x = 1/2019

Vậy x = 1/2018; 1/2019

\(2018x-1+2019x\left(1-2018x\right)=0\)

\(-\left(1-2018x\right)+2019x\left(1-2018x\right)=0\)

\(\left(1-2018x\right)\left(-1+2019x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}1-2018x=0\\-1+2019x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2018}\\x=\frac{1}{2019}\end{cases}}}\)

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

pkm;lkml

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 \(x=2017\Rightarrow A=2017^4.0-2017^3.0+2017^2.0-2017.0+2017-1000\)

\(A=2017-1000=1017\)