Vì f(0)=4 => c=4

=> f(x)=ax^2+bx+4

Vì f(1)=3 => a+b+4=3 => a+b=-1(1)

f(-1)=7 => a-b+4=7 => a-b =3 (2)

Từ (1),(2) => a = 1; b=-2 

=> f(x)=x^2-2x+4

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

Cộng vế với vế \(a+b+a-b=2013+2012\Leftrightarrow2a=4025\Leftrightarrow a=\frac{4025}{2}\)

\(\Rightarrow b=\frac{4025}{2}-2012=\frac{1}{2}\)

Vậy \(a=\frac{4025}{2};b=\frac{1}{2};c=0\)

a: f(0)=3

=>\(a\cdot0+b=3\)

=>b=3

=>f(x)=ax+3

f(2)=7

=>2a+3=7

=>2a=4

=>a=2

=>f(x)=2x+3

b: f(2)=8

=>\(a\cdot2+b=8\)

=>b=-2a+8

f(-2)=12

=>\(a\cdot\left(-2\right)+b=12\)

=>-2a+b=12

=>-2a-2a+8=12

=>-4a=4

=>a=-1

=>\(b=-2a+8=-2\cdot\left(-1\right)+8=2+8=10\)

Vậy: f(x)=-x+10

f(-7)=7

=>a*(-7)^2021+b*(-7)^2019+c*(-7)-5=7

=>a*7^2021+b*7^2019+c*7+5=-7

=>f(7)+10=-7

=>f(7)=-17

\(f\left(0\right)=ax^2+bx+c=a.0^2+b.0+c=c=4\)

\(f\left(1\right)=ax^2+bx+c=a+b+c=3\)

\(f\left(-1\right)=a-b+c=7\)

Ta có hpt \(\hept{\begin{cases}c=4\\a+b+c=3\\a-b+c=7\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=-1\left(1\right)\\a-b=3\left(2\right)\end{cases}}\)

Lấy (1) - (2) ta được : \(2b=-4\Rightarrow b=-2\)

Thay b = -2 vào (1) \(a-2=-1\Rightarrow a=1\)

Vậy \(\left(a;b;c\right)=\left(1;-2;4\right)\)

Ta có : \(\left\{{}\begin{matrix}f\left(a\right)=2a^2+b=0\\f\left(b\right)=b^2+ab+b=0\\2a^2=b^2+ab\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a^2+b=0\\a+b=-1\\a^2-b^2=\left(a+b\right)\left(a-b\right)=ab-a^2=a\left(b-a\right)\end{matrix}\right.\)

\(\Rightarrow\left(a+b\right)\left(a-b\right)+a\left(a-b\right)=\left(a-b\right)\left(2a+b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=b\\a+b=-a=-1\end{matrix}\right.\)

TH1 : a = b .

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

TH2 : a = 1

\(\Rightarrow b=-2\)