\(f\left(0\right)=a.0^2+b.0+c=c\) có giá trị nguyên 

\(f\left(1\right)=a+b+c\) có giá trị nguyên => a + b có giá trị nguyên 

\(f\left(2\right)=4a+2b+c=2a+2\left(a+b\right)+c\)=> 2a có giá trị nguyên 

=> 4a có giá trị nguyên 

=> 2b có giá trị nguyên.

Ta có:

\(f\left(0\right)=c\in Z\)(1)

\(f\left(1\right)=a+b+c\in Z\)(2)

\(f\left(2\right)=4a+2b+c\in Z\)(3)_

Từ (1), (2) => \(a+b\in Z\)=> \(2a+2b\in Z\)(4)

Từ (1), (3)=> 4a+2b\(\in Z\)(5)

Từ (4), (5) => \(\left(4a+2b\right)-\left(2a+2b\right)\in Z\)

=> \(2a\in Z\)=> \(2b\in Z\)

 chịu 

Quá dễ

13a+b+2c=0

=>b=-13a-2c

f(-2)=4a-2b+c=4a+c+26a+4c=30a+5c

f(3)=9a+3b+c=9a+c-39a-6c=-30a-5c

=>f(-2)*f(3)<=0

Tham khảo:

Có  \(c=2a+4b\). Ta tính f ( -1 ) và f ( 2 )

\(f\left(-1\right)=a-b+c=a-b+2a+4b=3a+3b=3\left(a+b\right)\)

\(f\left(2\right)=4a+2b+c=4a+2b+2a+4b=6a+6b=6\left(a+b\right)\)

\(\Rightarrow f\left(-1\right).f\left(2\right)=3\left(a+b\right).6\left(a+b\right)=18\left(a+b\right)^2\)

Có \(\left(a+b\right)^2\ge0\forall x\Leftrightarrow18\left(a+b\right)^2\ge0\forall x\left(đpcm\right)\)

Lời giải:

Có \(f(0),f(1),f(2)\in\mathbb{Z}\Rightarrow \left\{\begin{matrix} f(0)=c\in\mathbb{Z}\\ f(1)=a+b+c\in\mathbb{Z}\\ f(2)=4a+2b+c\in\mathbb{Z}\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} a+b\in\mathbb{Z}\\ 4a+2b\in\mathbb{Z}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2a+2b\in\mathbb{Z}\\ 4a+2b\in\mathbb{Z}\end{matrix}\right.\Rightarrow 2a\in\mathbb{Z}\rightarrow 2b\in\mathbb{Z}\)

Ta có đpcm