\(x^{2018}+2x^{2017}+3x^{2016}+...+2017x+2018\)

\(=1+2+3+...+2017+2018\)

\(=\frac{2018.\left(2018+1\right)}{2}=2037171\)

Lời giải:

Tại $x=2016$ thì $x-2016=0$

Khi đó:
$A=x^{2016}(x-2016)-x^{2015}(x-2016)+x^{2014}(x-2016)-x^{2013}(x-2016)+.....-x(x-2016)+x-2017$

$=x^{2016}.0-x^{2015}.0+......-x.0+2016-2017=2016-2017=-1$

2017 = 2016 + 1 = x + 1

suy ra 2017x¹⁵ = x¹⁶ + x¹⁵

2017x¹⁴ = x¹⁵ + x¹⁴

.... 

từ đó ta dễ tính ra A

Ta có:

   \(x^6-2017x^5+2017x^4-2017x^3+2017x^2-2017x+2017\)

\(=x^6-2016x^5-x^5+2016x^4+x^4-2016x^3-x^3+2016x^2+x^2-2016x-x+2017\)

\(=x^5\left(x-2016\right)-x^4\left(x-2016\right)+x^3\left(x-2016\right)-x^2\left(x-2016\right)+x\left(x-2016\right)-\left(x-2016\right)+1\)

Thay x = 2016 vào ta được giá trị biểu thức trên bằng 1

\(x^6-2017x^5+2017x^4-2017x^3+2017x^2-2017x+2017\) (1)

Thay 2017 = x+1 vào  (1) ,có :

\(x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\) 

= \(x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)  

= 1

f(2016)=2016^8 - 2017*2016^7 +2017*2016^6 - 2017*2016^5 +...+2017*2016^2 - 2017*2016+ 2018

=2016^8 -( 2016^8 + 2016) + (2016^7+2016) - (2016^6 + 2016)+....+2016^3+2016 -( 2016^2 + 2016)+2018

=2018

mình đọc chả hiểu gì 

có bạn nào giải chi tiết ra được không

A=x⁸-2016x⁷-x⁷+2016x⁶+x⁶-2015x⁵+....+2017x+x-2016+2018

  =x⁷(x-2016)-x⁶(x-2016)+.....-x(x-2016)+(x-2016)+2018

Do x=2016\(\Leftrightarrow\)x-2016=0

A=2018

Dễ thầy 2017=2016+1=x+1

Thay vào ta có:

\(x^{10}-2017x^9+2017x^8-.....+2017x^2-2017x+2017\)

\(=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-....+\left(x+1\right)x^2-\left(x+1\right)x+2017\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-....+x^3+x^2-x^2-x+2017=-x+2017=-2016+2017=1\)

Vậy..........

thanks bn!!

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