Xét đa thức \(F\left(x\right)=ax^2+bx+c\)

\(F\left(0\right)=c=2016\)

\(F\left(1\right)=a+b+c=2017\Rightarrow a+b=1\)  (1)

\(F\left(-1\right)=a-b+c=2018\Rightarrow a-b=2\)  (2)

Từ (1), (2)

\(\Rightarrow\hept{\begin{cases}a+b-a+b=-1\\a+b+a-b=3\end{cases}}\Rightarrow\hept{\begin{cases}2b=-1\\2a=3\end{cases}}\Rightarrow\hept{\begin{cases}b=-0,5\\a=1,5\end{cases}}\)

\(\Rightarrow F\left(2\right)=1,5.2^2-0,5.2+2016=2021\)

Vậy \(F\left(2\right)=2021\).

f(x) = ax^2 + bx + cf(1) = a + b + cf(-1) = a - b - cVì f(1) = f(-1) => a + b + c = a - b - c=> b = -b=> b = 0Vậy f(x) =  ax^2 + bx + c = ax^2 + cf(-x) = a(-x)^2 + 0 + c = ax^2 + c=> f(x) = f(-x)

Có : \(f\left(-1\right)=f\left(1\right)\)

\(\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=a.1^2+b.1+c\)

\(\Leftrightarrow a-b+c=a+b+c\)

\(\Leftrightarrow b=0\)

Khi đó \(f\left(x\right)=\) \(a.x^2+c\) và \(f\left(-x\right)=a.\left(-x\right)^2+c=a.x^2+c\)

Do vậy \(f\left(x\right)=f\left(-x\right)\)

Ta có : \(f\left(0\right)=c=1\)

\(f\left(1\right)=a+b+c=2\)

\(f\left(2\right)=4a+2b+c=8\)

\(\Rightarrow c=1,a=\frac{5}{2},b=\frac{-3}{2}\)

Vì vậy mà \(f\left(x\right)=\frac{5}{2}x^2-\frac{3}{2}x+1\)

nên \(f\left(-2\right)=\frac{5}{2}.\left(-2\right)^2-\frac{3}{2}.\left(-2\right)+1=14\)

Theo de ta co: 

   1)        a.x²+b.x+c = -2    . Thay x=0 vao bieu thuc nay duoc:

a.0² + b.0 + c = -2

=> 0+0+c = -2

=> c=-2

2)           a.x² + b.x+c = 1    . Thay x=1 vao bieu thuc nay duoc:

a.1² + b.1 + c = 1

=> a+ b + c = 1

Thay c=-2  vua tim o (1) vaobieu thuc tren duoc:

a+b-2 =1 => a+b =3

2)           a.x² +b.x +c = 4      . Thay x=-2 vao bieu thuc nay, ta duoc:

a.(-2)² + b.(-2) + c = 4

=> 4a - 2b + c = 4 

=> 2 ( 2a - b ) +c = 4

Thay: c = -2 tim o (1) vao bieu thuc tren duoc:

2(2a-b) -2 = 4 

=> 2a- b = (4+2):2 = 3

Bay gio ta da co 2 yeu to: 

a+b = 3            ;           2a-b = 3

Tu a+b = 3  => b = 3-a   . Thay b=3-a vao bieeu thuc 2a-b = 3 ta duoc:

2a - (3-a) = 3   => 2a - 3 + a = 3 => 3a = 3+3 =6  => a = 6:3 = 2

Suuy ra: b = 3-2 = 1

Vay: a=2   ;   b=1   ;   c = -2