\(a.f\left(1\right)=f\left(-1\right)\Leftrightarrow a+b+c=a-b+c\Leftrightarrow2b=0\Leftrightarrow b=0\)

\(\Rightarrow f\left(x\right)=ax^2+c\)

Khi đó ta có:

\(\left\{{}\begin{matrix}f\left(m\right)=am^2+c\\f\left(-m\right)=am^2+c\end{matrix}\right.\Rightarrow f\left(m\right)=f\left(-m\right)\forall m\)

Lời giải:

$f(1)=a+b+c=6$

$f(2)=4a+2b+c=16$

$f(12)-f(-9)=(144a+12b+c)-(81a-9b+c)$

$=63a+21b=21(3a+b)$

$=21[(4a+2b+c)-(a+b+c)]=21(16-6)=21.10=210$

\(f\left(0\right)=c⋮3\) ;

 \(f\left(1\right)=a+b+c⋮3\) mà \(c⋮3\Rightarrow a+b⋮3\)

\(f\left(-1\right)=a-b+c=-2b+\left(a+b+c\right)⋮3\)  mà \(a+b+c⋮3\Rightarrow-2b⋮3\Rightarrow b⋮3\) (do 2 và 3 nguyên tố cùng nhau)

\(\left\{{}\begin{matrix}a+b+c⋮3\\b⋮3\\c⋮3\end{matrix}\right.\) \(\Rightarrow a⋮3\)

a: f(1)=a+b+c=0

=>x=1 là nghiệm

b: Vì 5-6+1=0

nên f(x)=5x^2-6x+1 có một nghiệm là x=1

f(1)=f(-1)

=>\(a\cdot1^2+b\cdot1+c=a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c\)

=>a+b+c=a-b+c

=>2b=0

=>b=0

=>\(f\left(x\right)=ax^2+c\)

\(f\left(-x\right)=a\cdot\left(-x\right)^2+c=a\cdot x^2+c\)

=>f(x)=f(-x)

=>f(2021)=f(-2021)

=>f(2021)=2021

Mình giải giúp bạn nha

Giải :

Ta có : \(\int\left(x\right)=ãx^2+bx+c\)

\(\Rightarrow\int\left(-2\right)=4a-2b+c\) = 2a - 2b +2a + c = 2a -2b +3c +6 = 0

\(\Rightarrow2a-2b+3c=-6\) (1)

\(\int\left(2\right)=4a+2b+c\) = 2a + 2b + 2a + c = 2a + 2b +3c +6 =0

\(\Rightarrow2a+2b+3c=-6\) (2)

Từ (1) và (2) \(\Rightarrow2a-2b+3c=2a+2b+3c\)

\(\Rightarrow2a-2b+3c-\left(2a+2b+3c\right)=0\)

\(\Rightarrow-4b=0\)

\(\Rightarrow b=0\)

\(\Rightarrow2a+3c=-6\)

\(\Rightarrow5c+6=-6\)

\(\Rightarrow5c=-12\)

\(\Rightarrow c=\dfrac{-12}{5}\)

\(\Rightarrow a=\dfrac{-12}{5}+3\)

\(\Rightarrow a=\dfrac{3}{5}\)

Vậy \(b=0;c=\dfrac{-12}{5};a=\dfrac{3}{5}\)

Ta có : a + c = b + 2018

b = a + c - 2018

f(-1) =  a . ( -1 )² + b . ( -1 ) + c = a - b + c = a - ( a + c - 2018 ) + c = a - a - c + 2018 + c = 2018

f(0)=2018

=>\(a\cdot0^2+b\cdot0+c=2018\)

=>c=2018

=>f(x)=\(ax^2+bx+2018\)

f(1)=2019

=>\(a\cdot1^2+b\cdot1+2018=2019\)

=>a+b=1

f(-1)=2020

=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+2018=2020\)

=>a-b=2

mà a+b=1

nên a=(2+1)/2=1,5; b=a-2=1,5-2=-0,5

=>f(x)=1,5x^2-0,5x+2018

f(2)=1,5*2^2-0,5*2+2018

=6-1+2018

=2018+5

=2023