Bài làm:

Vì x=2011 => x+1=2012(*)

Thay (*) vào f(x) ta được:

f(2011) = x⁶ - (x+1)x⁵ + (x+1)x⁴ - (x+1)x³ + (x+1)x² - (x+1)x + 2017

f(2011) = x⁶ - x⁵ - x⁴ + x³ + x² - x² - x +2017

f(2011) = -x +2017

f(2011) = -2011 + 2017

f(2011) = 6

Học tốt!!!!

Ta có: 2012=2011+1=x+1

\(A=x^5-2012x^4+2012x^3-2012x^2+2012x-2012\\ =x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-\left(x+1\right)\\ =x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x-1\\ =-1\)

Thay 2012 = x + 1

\(B=x^{2011}-\left(x+1\right).x^{2010}+\left(x+1\right).x^{2009}+...-\left(x+1\right).x^2+\left(x+1\right).x-1\)

\(=x^{2011}-x^{2011}-x^{2010}+x^{2010}+x^{2009}-...-x^3-x^2+x^2+x-1\)

\(=x-1=2011-1=2010\)

Ta có : \(x=2011\Rightarrow x+1=2012\)

Khi đó :

\(x^{10}-2012x^9+2012x^8-2012x^7+....-2012x+2012\)

\(=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...-\left(x+1\right)x+x+1\)

\(=x^{10}-x^9-x^8+x^8+x^7-x^7-x^6+...-x^2-x+x+1\)

\(=1\)

\(x=2011\Rightarrow2012=x+1\)

\(\Rightarrow M\left(2011\right)=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+1\)

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

\(=-x+1=-2011+1=-2010\)

Với x = 2011 => x + 1 = 2012

=> A = x¹⁰ - ( x + 1 )x⁹ + ( x + 1)x⁸ - ( x+ 1)x⁷ + ( x + 1 )x⁶ - ( x + 1 )x⁵+ ( x + 1 )x⁴ - ( x + 1 )x³ + ( x + 1)x² - ( x + 1 )x + 2012

        = x¹⁰ - x¹⁰ - x⁹ + x⁹ + x⁸ - x⁸ - x⁷ + x⁷+ x⁶- x⁶ - x⁵ + x⁵ + x⁴ - x⁴ - x³ + x³ + x² - x² - x + 2012

        = -x  + 2012

Thay x=2011 vào ta được: ( - 2011 ) + 2012 = 1

Giải:

Thay \(2012=x+1\) vào biểu thức ta có:

\(\Rightarrow B=x^{2011}-\left(x+1\right).x^{2010}+\left(x+1\right).x^{2009}-...-\left(x+1\right).x^2+\left(x+1\right).x-1\)

\(=x^{2011}-x^{2011}-x^{2010}+x^{2010}+x^{2009}-...-x^2+x^2+x-1\)

\(=x-1\)

\(\Rightarrow B=2011-1=2010\)

Vậy \(B=2010\)