thay x=2014 vào ta có:

f(2014)=2014²⁰¹⁴-2015.2014²⁰¹³+2015.2014²⁰¹²-2015.2014²⁰¹¹+...-2015.2014+2015

=2014²⁰¹⁴-(2014+1)2014²⁰¹³+(2014+1).2014²⁰¹²-(2014+1).2014²⁰¹¹+...-(2014+1).2014+2014+1

=2014²⁰¹⁴-2014²⁰¹⁴-2014²⁰¹³+2014²⁰¹³+2014²⁰¹²-2014²⁰¹²-2014²⁰¹¹+...-2014²-2014+2014+1

=1

Bạn ko thấy là cái đề thiếu trầm trọng à
 

bố ông ko ghi đề thì làm sao mà làm

x = 2014 => x + 1 = 2015

=> f(2014) = x²⁰¹⁴ - (x + 1).x²⁰¹³ + (x + 1).x²⁰¹² - ... - (x + 1).x + x + 1

= x²⁰¹⁴ - x²⁰¹⁴ - x²⁰¹³ + x²⁰¹³ + x²⁰¹² - ... - x² - x + x + 1

= 1

minh moi hoc lop 5

=> \(f\left(x\right)=x^{2014}-\left(2014+1\right)x^{2013}+\left(2014+1\right)x^{2012}+...-\left(2014+1\right)x+2014+1\)

Mà x = 2014

=> \(f\left(2014\right)=x^{2014}-\left(x+1\right)x^{2013}+\left(x+1\right)^{2012}+...-\left(x+1\right)x+x+1\)

\(=x^{2014}-x^{2014}+x^{2013}-x^{2013}-x^{2012}+....-x^2-x+x+1\)

\(=1\)

=> f(2014) = 1

thank nha

Với x = 2014 => x + 1 = 2015

Ta có : 

\(f\left(x\right)=x^{17}-2015x^{16}+2015x^{15}-2015x^{14}+....+2015x-1\)

<=>  \(f\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-\left(x+1\right)x^{14}+....+\left(x+1\right)x-1\)

<=> \(f\left(x\right)=x-1\)

<=> \(f\left(2014\right)=2014-1=2013\)