f(2015)=a(2015)^5+b(2015)^3+2014.2015 +1 mà f(2015)=2 => a(2015)^5+b(2015)^3+2014.2015+1=2 =>a(2015)^5+b(2015)^3+2014.2015 =1
Xét f(-2015)=a(-2015)^5+b(-2015)^3+2014.(-2015) +1=-a(2015)^5-b(2015)^3-2014.2015 +1 = -(a(2015)^5+b(2015)^3+2014.2015)+1 =-1+1=0
bài dễ
ta có f(2015)=a.2015^5+b.2015^3+2014.2015+1
f(-2015)=a.(-2015)^5+b.(-2015)^3+2014.(-2015)+1
=>f(2015)+f(-2015)=2
(=)2+f(-2015)+2
(=) f(-2015)=0
Ta có:\(f\left(x\right)=ax^5+bx^3+2014x+1\)
\(f\left(-x\right)=-ax^5-bx^3-2014x+1\)
\(\Rightarrow f\left(x\right)+f\left(-x\right)=\left(ax^5+bx^3+2014x+1\right)-\left(x^5+bx^3+2014x-1\right)\)
\(=2\)
\(\Rightarrow f\left(2015\right)+f\left(-2015\right)=2\)
Mà \(f\left(2015\right)=2\Rightarrow f\left(-2015\right)=0\)