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

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

\(Q\left(2\right)+Q\left(-1\right)=5a+b+2c=0\)

\(\Leftrightarrow Q\left(2\right)=-Q\left(-1\right)\)

\(Q\left(2\right).Q\left(-1\right)=-Q\left(-1\right)^2\le0\)

P(1)=a+b+c

P(-2)=4a-2b+c

P(1)+P(2)=5a-3b+2c=0 => P(1) và P(2) trái dấu hoặc P(1)=P(2)=0

=>p(1).P(2) bé hơn hoặc bằng không

Ta có: P(x)=ax² + bx + c.

=> P(1)= a.1²+b.1+c=a+b+c.

     P(-2)=a.(-2)²+b.(-1)+c=4a-2b+c.

Ta lại có: P(1)+P(-2)= (a+b+c)+(4a-2b+c)=5a-b+2c=0.

=> P(1)= -P(-2).

=> P(1).P(-2)= -P(-2).P(-2)= - [ P(-2)]²  <_{_} 0.

Vậy: P(1).P(-2)<_{_} 0

Bài làm:

Ta có: \(Q\left(2\right)=4a+2b+c\)

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

\(\Rightarrow Q\left(2\right)+Q\left(-1\right)=5a+b+2c=0\)

\(\Rightarrow Q\left(2\right)=-Q\left(-1\right)\)

Ta có: \(Q\left(2\right).Q\left(-1\right)=-Q\left(-1\right).Q\left(-1\right)=-Q\left(-1\right)^2\le0\)

=> đpcm

Học tốt!!!!

Cũng có thể sai =)

Có :

\(P\left(-1\right)=a-b+c\le\)\(P\left(1\right)=a+b+c\le P\left(2\right)=4a+2b+c\)

\(P\left(1\right)+P\left(2\right)=5a+2b+2c=0\)

\(\Rightarrow P\left(2\right)=-P\left(1\right)\)

\(\Rightarrow P\left(2\right).P\left(1\right)\le0\)

Mà \(P\left(-1\right)\le P\left(1\right)\)

\(\Rightarrow P\left(-1\right).P\left(2\right)\le0\)

\(\Delta\) = b² - 4ac = (5a + 2c)² - 4ac = 25a² + 20ac + 4c² - 4ac = 25a² + 16ac + 4c² 

= 9a² + (16a² + 16ac + 4c²)

= 9a² + (4a + 2c)² \(\ge\) 0 với mọi a; c

=> Pt đã cho luôn có nghiệm

Thay `b=5a+2c` vào `ax^2+bx+c=0`:

`ax^2+(5a+2c)x+c=0`

`=>Delta=(5a+2c)^2-4ac`

`=25a^2+20ac+4c^2-4ac`

`=25a^2+16ac+4c^2`

`=9a^2+(16a^2+16ac+4c^2)`

`=9a^2+(4a+2c)^2>=0`

`=>` ĐPCM