- Xét x^8 - 2014x^7 tại x= 2013

   x^8 - 2014x^7 = 2013^8-2014. 2013^7 = 2013^7. ( 2013-2014) = - 2013^7

- Tính tiếp : -2013^7 + 2014. 2013^6 = 2013^6 ( -2013+2014 ) = 2013^6

Cứ như vậy đến -2014. 2013 = - 2013

Cuối cùng KQ f(2013) = -2013+2020=7

2013^8 - 2014. 2013^7 = 2013^7 ( 2013-2014 ) = - 2013^7

- 2013^7 + 2014. 2013^6 = 2013^6 ( -2013+2014) = 2013^6

..............................................................................................

..............................................................................................

-2013 + 2020 = 7

Vậy f(2013)= 7

ta có f(x)=ax⁵+bx³+2014x+1 \(\Rightarrow\)f(2015)=2015⁵a+2015³b+2014.2015+1 và f(-2015)=(-2015)⁵a+(-2015)³b+2014(-2015)+1

\(\Rightarrow\)f(2015)+f(-2015)=2015⁵a+2015³b+2014.2015+1+(-2015)⁵a+(-2015)³b+2014(-2015)+1=2

\(\Rightarrow\)f(2015)+f(-2015)=2 mà f(2015)=2 \(\Rightarrow\)f(-2015)=0 vậy............

NHỚ K CHO ME NHÉ !!!

olm chu ko phai la onl nhe

lên google mà kham khảo

ok còn ai ko 

câu a

ta có \(\hept{\begin{cases}f\left(0\right)=c=0\\f\left(1\right)=a+b+c=2013\\f\left(-1\right)=a-b+c=2012\end{cases}\Rightarrow\hept{\begin{cases}c=0\\a=2012,5\\b=0,5\end{cases}}}\)

câu b , do \(f\left(-2\right)=f\left(3\right)\Leftrightarrow4a-2b+c=9a+3b+c=2036\)

\(f\left(1\right)=a+b+c=2012\Rightarrow\hept{\begin{cases}a=4\\b=-4\\c=2012\end{cases}}\)do đó \(f\left(x\right)=4x^2-4x+2012=\left(2x-1\right)^2+2011>0\)với mọi x,

Ta có: a = 2020 => 2021 = x + 1

f(2020) = x²⁰¹⁴ - (x + 1) . x²⁰¹³ + (x + 1) . x²⁰¹² - ... + (x + 1) . x² - (x + 1) . x - 1

= x²⁰¹⁴ - x²⁰¹⁴ + x²⁰¹³ + x²⁰¹³ + x²⁰¹² - ... + x³ + x² - x² + x - 1

= x - 1 = 2020 - 1 = 2019

Vậy f(2020) = 2019

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