x=2014 => x+1 = 2015

f(2014) = x^17 - (x+1)x^16 + ... + (x+1)x -1
= x^17 - x^17 - x^16 + x^16 - x^15 - ... + x^2 + x -1
= x - 1 = 2013

Ta thấy \(x=2014\Rightarrow x+1=2015\)

Ta có: \(f\left(2014\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-...+\left(x+1\right)x-1\)

                     \(=x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-...+x^2+x-1\)

                     \(=x-1\)(1)

Thay x=2014 vào  (1) ta được:

   \(f\left(2014\right)=2014-1\)

                      \(=2013\)

Ta có :\(x=2014\Rightarrow2015=x+1\)

\(\Rightarrow f\left(2014\right)=x^{17}-\left(x+1\right)x^{2016}+\left(x+1\right)x^{2015}-.....+\left(x+1\right)x-1\)

\(=x^{17}-x^{17}-x^{2016}+x^{2016}+x^{2015}-....+x^2+x-1\)

\(=x-1=2014-1=2013\)

Cảm ơn bạn nhiều !

ở cuối có 1 số 1 thôi các bạn nhé!

tính f(2014) nha

f(x) = x¹⁷-2015x¹⁶+2015x¹⁵-2015x¹⁴+...+2015x-1

ta có x=2014

=> 2015=2014+1=x+1

f(x)=x¹⁷-(x+1)x¹⁶+(x+1)x¹⁵-(x+1)x¹⁴+...+(x+1)x-1

=x¹⁷-x¹⁷-x¹⁶+x¹⁶+x¹⁵-x¹⁵-x¹⁴+...+x²+x-1

=x-1

=2014-1=2013

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

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