\(P\left(2\right)=4a+2b+c=2\left(5a+b+2c\right)-6a-3c=-6a-3c\)

\(P\left(-1\right)=a-b+c=-\left(5a+b+2c\right)+6a+3c\)

\(\Rightarrow P\left(2\right).P\left(-1\right)=\left(-6a-3c\right)\left(6a+3c\right)=-\left(6a+3c\right)^2\le0\) (đpcm)

\(a=1,b=6,c=1\)

\(5a-b+c=5-6+1=0\)

\(P\left(1\right).P\left(3\right)=\left(1.1^2+6.1+1\right).\left(1.3^2+6.3+1\right)>0?\)

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.

Sửa đề: CMR: \(P\left(-1\right).P\left(-2\right)\le0\)

Ta có:

\(P\left(x\right)=ax^2+bx+c\)

\(\Rightarrow\left\{{}\begin{matrix}P\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c\\P\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}P\left(-1\right)=a-b+c\\P\left(-2\right)=4a-2b+c\end{matrix}\right.\)

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

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

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

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

Mà \(P^2\left(-2\right)\ge0\Leftrightarrow-P^2\left(-2\right)\le0\)

Vậy \(P\left(-1\right).P\left(-2\right)\le0\) (Đpcm)

Ta có:

P(1)=\(a.1^2+b.1+c=a+b+c\) (1)

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

Từ (1) và (2) \(\Rightarrow P\left(1\right)+P\left(-2\right)=\left(a+b+c\right)+\left(4a-2b+c\right)\)

\(=a+b+c+4a-2b+c=5a-b+2c=0\) (theo đề bài)

Do P(1)+P(-2)=0 nên P(1) và P(-2) trái dấu \(\Rightarrow P\left(1\right).P\left(-2\right)\le0\)

Vậy...

vì P(x)=ax^2+bx+c

suy ra: p(1)=a.1^2+b1+c=a+b+c

p(-2)=a.(-2)^2-b2+c=(a5-b+2c)-(a+b+c)=0-(a+b+c)=-(a+b+c)

suy ra : p(1).p(-2)=(a+b+c).-(a+b+c)=-(a+b+c)^2

vì (a+b+c)^2>hoặc=0

suy ra -(a+b+c)^2<hoặc =0

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

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

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

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

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