\(F\left(x\right)-F\left(x-1\right)=x\)

\(\Leftrightarrow ax^2+bx-a\left(x-1\right)^2-b\left(x-1\right)=x\)

\(\Leftrightarrow2ax-a+b=x\)

Đồng nhất hệ số 2 vế:

\(\Rightarrow\left\{{}\begin{matrix}2a=1\\-a+b=0\end{matrix}\right.\) \(\Rightarrow a=b=\dfrac{1}{2}\)

Em cám ơn ah

Lời giải:

Ta có:
$f(1)=a+b+c$
$f(-2)=4a-2b+c$

$\Rightarrow 2f(-2)+3f(1)=2(4a-2b+c)+3(a+b+c)=11a-b+5c=0$

$\Rightarrow f(-2)=\frac{-3}{2}f(1)$

Vì $\frac{-3}{2}<0$ nên $f(-2)$ và $f(1)$ không thể cùng dấu.

f(2)=156

=>\(a\cdot2^2+b\cdot2+c=156\)

=>4a+2b+c=156

f(-3)=156

=>\(a\cdot\left(-3\right)^2+b\cdot\left(-3\right)+c=156\)

=>9a-3b+c=156

f(-1)=132

=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c=132\)

=>a-b+c=132

9a-3b+c=156

4a+2b+c=156

Do đó: 9a-3b+c-4a-2b-c=156-156

=>5a-5b=0

=>a=b

9a-3b+c=156

a-b+c=132

Do đó: 9a-3b+c-a+b-c=156-132

=>8a-2b=24

=>8a-2a=24

=>6a=24

=>a=4

=>b=a=4

a-b+c=132

=>4-4+c=132

=>c=132

\(f\left(x\right)=4x^2+4x+132\)

\(=4x^2+4x+1+131=\left(2x+1\right)^2+131\ge131>0\forall x\)

=>f(x) vô nghiệm

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

*Thay x=1=>f(1)+f(-1)=1+1=2

*Thay x=-1=>f(-1)-f(1)=-1+1=0

=>f(1)+f(-1)-(f(-1)-f(1))=2-0

=>2.f(1)=2

=>f(1)=1

f(1) + 1.f(-1) = 1+ 1 = 2 => f(1) + f(-1) = 2  (*)

f(-1) + (-1). f(1) = -1 + 1 = 0 => f(-1) - f(1) = 0 => f(-1) = f(1). Thay vào (*)

=> 2. f(1) = 2 => f(1) = 1