bạn dùng phương pháp hệ số bất định là ra mà

Ta có: P(-1).P(-2)=[a.(-1)²+b.(-1)+c].[a.(-2)²+b.(-2)+c]

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

=[(5a-4a)-(3b-2b)+(2c-c)].(4a-2b+c)

=(5a-4a-3b+2b+2c-c).(4a-2b+c)

=[(5a-3b+2c)-(4a-2b+c)].(4a-2b+c)

Vì 5a-3b+2c=0

=>P(-1).P(-2)=[0-(4a-2b+c)].(4a-2b+c)

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

=-(4a-2b+c)² 

Vì \(\left(4a-2b+c\right)^2\ge0\)

=>\(-\left(4a-2b+c\right)^2\le0\)

=>\(P\left(-1\right).P\left(-2\right)\le0\)

=>ĐPCM

Lê Chí Cường lm dài thế

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

Lời giải:
$C(2)=a.2^2+b.2+c=4a+2b+c$
$C(-1)=a(-1)^2+b(-1)+c=a-b+c$

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

$\Rightarrow C(-1)=-C(2)$

$\Rightarrow C(2)C(-1)=-C(2)^2\leq 0$ 

Ta có đpcm.

\(\Delta\) = 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

a) \(P\left(-1\right)=a-b+c\)

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

b) \(P\left(-1\right)+P\left(-2\right)=5a-3b+2c=0\)

=> P ( - 1) = -P(-2) 

=> P( -1 ) . P (-2) \(=-\left[P\left(-2\right)\right]^2\le0\)

a) \(\text{P}\left(-1\right)=\text{a}+\text{b}+\text{c}\)

\(\text{P}\left(-2\right)=4\text{a}-2\text{b}+\text{c} \)

b) \(\text{P}\left(-1\right)+\text{P}\left(-2\right)=5\text{a}+3\text{b}+2\text{c}=0\)

\(\Rightarrow\text{ P}\left(-1\right)=\text{P}\left(-2\right)\)

\(\Rightarrow\text{ P}\left(-1\right).\text{ P}\left(-2\right)=\left[\text{P}\left(-2\right)\right]^2\le0\)