f(0) = 2020
=> a.02 + b.0 + c = 2020
=> c = 2020
F(1) = 2021
=> a.12 + b1 + c = 2021
=> a + b + 2020 = 2021 (Vì c = 2020)
=> a + b = 1 (1)
F(-1) = 2019
=> a.(-1)2 + b.(-1) + c = 2019
=> a - b + 2020 = 2019
=> a - b = -1 (2)
Từ (1)(2) => a = 0 ; b = 1
=> f(x) = x + 2020
=> f(2022) = 2022 + 2020 = 4042
Ta có : \(f\left(0\right)=2020< =>a.0+b.0+c=2020< =>c=2020\)(1)
\(f\left(1\right)=2021< =>a+b+c=2021< =>a+b=1\)(2)
\(f\left(-1\right)=2019< =>a-b+c=2019< =>a-b=-1\)(3)
Từ (1) ; (2) và (3) \(=>\hept{\begin{cases}a+b=1\\a-b=-1\\c=2020\end{cases}}< =>\hept{\begin{cases}a=0\\b=1\\c=2020\end{cases}}\)
Suy ra \(f\left(2022\right)=2022^2.0+2022.1+2020=4042\)
