Question

[Giấc mơ trưa]

Cho đa thức f(x)=ax⁵+bx³+2014x+1, biết f(2015)=2. Hãy tính f(-2015)

Help me!!!!!!!!!!!!!

Answer 1

[zZz Cool Kid_new zZz]

\(f\left(x\right)=ax^5+bx^3+2014x+1\)

\(\Rightarrow f\left(-x\right)=a\left(-x\right)^5+b\left(-x\right)^3+2014\left(-x\right)+1\)

\(=-ax^5-bx^3-2014x+1\)

\(\Rightarrow f\left(x\right)+f\left(-x\right)=2\)

\(\Rightarrow f\left(2015\right)+f\left(-2015\right)=2\)

Mà \(f\left(2015\right)=2\Rightarrow f\left(-2015\right)=0\)

Answer 2

[Rinz Rinz]

f(-2015) = 0