đối xứng qua x=0

=> b=0

4a+c=0

c=-4a

c<a=> -4a<a => a>0 

chưa thể xác định chính xác đươc

đa thức là  ax^2-4a=0 với a>0

Ta có:

\(f\left(0\right)=a.0^2+b.0+c=0\)

\(=0+0+c=0\Rightarrow c=0\)

\(f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=0\)

\(a-b+0=0\)

\(\Rightarrow a-b=0\)

\(\Rightarrow a=b\)

\(f\left(1\right)=a.1^2+b.1+c=0\)

\(\Rightarrow a+b+0=0\)

\(\Rightarrow a+b=0\)

Mà \(a=b\)

\(\Rightarrow a=b=\frac{0}{2}=0\)

Vậy \(a=b=c=0\)

\(f\left(0\right)=2010\Rightarrow a.0^2+b.0+c=2010\Rightarrow c=2010\)

\(f\left(1\right)=2011\Rightarrow a.1^2+b.1+c=2011\Rightarrow a+b+c=2011\)

\(\Rightarrow a+b+2010=2011\Rightarrow a+b=1\) (1)

\(f\left(-1\right)=2012\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=2012\)

\(\Rightarrow a-b+c=2012\Rightarrow a-b+2010=2012\)

\(\Rightarrow a-b=2\Rightarrow a=b+2\)

Thế vào (1) \(\Rightarrow b+2+b=1\Rightarrow2b=-1\Rightarrow b=-\dfrac{1}{2}\)

\(\Rightarrow a=b+2=-\dfrac{1}{2}+2=\dfrac{3}{2}\)

\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^2-\dfrac{1}{2}x+2010\)

\(\Rightarrow f\left(-2\right)=\dfrac{3}{2}.\left(-2\right)^2-\dfrac{1}{2}.\left(-2\right)+2010=2017\)

\(\left\{{}\begin{matrix}f\left(0\right)=5\Rightarrow0+0+5\Rightarrow c=5\\f\left(1\right)=0\Rightarrow a+b+5=0\\f\left(5\right)=0\Rightarrow25a+5b+5=0\end{matrix}\right.\) \(\left\{{}\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\end{matrix}\right.\)

tu (3) => b =-1-5a

tu (2) => a-1-5a+5 =0 => a =1 ;b =-6

y =x^2 -6x +5

y(-1) =1 +6 +5 khac 3 => loai

y(-1/2) =1/4 -6/2 +5 =1/4 +2 = 9/4 nhan

Q(1/2;9/4) thuoc dths

sky oi say oh yeah

Ta có: \(f\left(x\right)=ax^2+bx+c\)

\(\Rightarrow f\left(0\right)=c⋮3\Rightarrow c⋮3\)

\(\left\{{}\begin{matrix}f\left(1\right)=a+b+c⋮3\\f\left(-1\right)=a-b+c⋮3\end{matrix}\right.\)

Mà \(c⋮5\)

\(\Rightarrow\left\{{}\begin{matrix}a+b⋮3\\a-b⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a⋮3\\2b⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a⋮3\\b⋮3\end{matrix}\right.\) ( do \(\left(2;3\right)=1\) )

Vậy \(a,b,c⋮3\)