Question

Cho \(P\left(x\right)=\)\(x^4-2015x^3+2015x^2-2015x+2015\)Tính P(2014)

Answer 1

Nếu \(x=2014\Rightarrow x+1=2015\)

Ta có : 

\(P\left(x\right)=x^4-2015x^3+2015x^2-2015x+2015\)

\(\Rightarrow P\left(2014\right)=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(\Rightarrow P\left(2014\right)=x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(\Rightarrow P\left(2014\right)=0+0+0+0+1\)

\(\Rightarrow P\left(2014\right)=1\)

Vậy \(P\left(2014\right)=1\)