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

Vì \(x=2017\)

\(\Leftrightarrow x+1=2018\)

Thay vào P(x) ta được :

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

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

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

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

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

\(A=2017-1000=1017\)

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

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

tối nay mk phải nộp rùi

giúp mk vs!!

A=2018x-2017x-2016x-x

A=(2018-2017-2016-...-1)x

A=[(2018-2017)-(2017-2016)-....-(2-1)].x

A=(1-1-....-1)x

A=[1-(1+1+...+1)]x

A=(1-1008)x

A=-1007x

Thay x=2017 vào A ta có

A=-1007.2017= -2031119

Vậy A=-2031119

ĐK \(2018x\ge0\Rightarrow x\ge0\)

Khi đó \(x+\frac{1}{2018}\ge0;x+\frac{2}{2018}\ge0;...;x+\frac{2017}{2018}\ge0\)

Ta có \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)(Vế trái có 2017 hạng tử)

<=> \(x+\frac{1}{2018}+x+\frac{2}{2018}+...+x+\frac{2017}{2018}=2018x\)

<=> \(\left(x+x+...x\right)+\left(\frac{1}{2018}+\frac{2}{2018}+...+\frac{2017}{2018}\right)=2018x\)

           2017 hạng tử x                   2017 số hạng

=> \(2017x+\frac{1+2+...+2017}{2018}=2018x\)

=> \(x=\frac{2017.\left(2017+1\right):2}{2018}\)

\(\Rightarrow x=\frac{2017}{2}=1008,5\)(tm)

Vậy x = 1008,5

Vì \(\left|x+\frac{1}{2018}\right|\ge0\forall x\)

    \(\left|x+\frac{2}{2018}\right|\ge0\forall x\)

    \(\left|x+\frac{3}{2018}\right|\ge0\forall x\)

     .......................................

    \(\left|x+\frac{2017}{2018}\right|\ge0\forall x\)

\(\Rightarrow\)\(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|\ge0\forall x\)

mà \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)

 \(\Rightarrow\)\(2018x\ge0\forall x\)\(\Rightarrow\)\(x\ge0\)

 \(\Rightarrow\)\(x+\frac{1}{2018}+x+\frac{2}{2018}+x+\frac{3}{2018}+...+x+\frac{2017}{2018}=2018x\)

\(\Leftrightarrow\)\(2017x+\frac{1}{2018}+\frac{2}{2018}+\frac{3}{2018}+...+\frac{2017}{2018}=2018x\)

\(\Leftrightarrow\)\(\frac{1+2+3+...+2017}{2018}=x\)

\(\Leftrightarrow\)\(x=\frac{\left[\left(2017+1\right).2017\right]:2}{2018}\)

\(\Leftrightarrow\)\(x=\frac{2035153}{2018}\)

\(\Leftrightarrow\)\(x=\frac{2017}{2}=1008,5\)

Vậy \(x=1008,5\)