Ta có P(-1) = a - b + c

P(3) = 9a + 3b +c

=> P(3) - P(-1) = (9a + 3b + c) - ( a - b + c) = 8a + 4b

Mà 2a + b = 0 (GT) => 8a + 4b = 0 => P(3) - P(-1) = 0

=> P(3) = P(-1) => P(3). P(-1) = (P(3))^2 lớn hơn hoặc = 0 (đpcm)

ta có: 2a + b  = 0

\(\Rightarrow2a=-b\Rightarrow a=\frac{-b}{2}\)

ta có: \(P_{\left(-1\right)}=a.\left(-1\right)^2+b.\left(-1\right)+c\)

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

thay số: \(P_{\left(-1\right)}=\frac{-b}{2}-b+c\)

\(P_{\left(-1\right)}=\frac{-b}{2}-\frac{2b}{2}+c=\frac{-b-2b}{2}+c\)

\(P_{\left(-1\right)}=\frac{-3b}{2}+c\)

ta có: \(P_{\left(3\right)}=a.3^2+b.3+c\)

\(P_{\left(3\right)}=a9+3b+c\)

thay số: \(P_{\left(3\right)}=\frac{-b}{2}.9+3b+c\)

\(P_{\left(3\right)}=\frac{-9b}{2}+\frac{6b}{2}+c\)

\(P_{\left(3\right)}=\frac{-9b+6b}{2}+c\)

\(P_{\left(3\right)}=\frac{-3b}{2}+c\)

\(\Rightarrow P_{\left(-1\right)}.P_{\left(3\right)}=\left(\frac{-3b}{2}+c\right).\left(\frac{-3b}{2}+c\right)\)

\(P_{\left(-1\right)}.P_{\left(3\right)}=\left(\frac{-3b}{2}+c\right)^2\ge0\)

\(\Rightarrow P_{\left(-1\right)}.P_{\left(3\right)}\ge0\left(đpcm\right)\)

Ta có : 

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

\(\Rightarrow\hept{\begin{cases}P\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c\\P\left(3\right)=a.3^2+b.3+c\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}P\left(-1\right)=a-b+c\\P\left(3\right)=9a+3b+c\end{cases}}\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=\left(9a+3b+c\right)-\left(a-b+c\right)\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=9a+3b+c-a+b-c\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=8a+4b\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=4\left(2a+b\right)\)

Mà \(2a+b=0\Rightarrow4\left(2a+b\right)=0\Rightarrow P\left(3\right)-P\left(-1\right)=0\Rightarrow P\left(3\right)=P\left(-1\right)\)

Nên : 

\(P\left(3\right).P\left(-1\right)=P\left(-1\right).P\left(-1\right)=\left[P\left(-1\right)\right]^2\ge0\)

\(\Rightarrow P\left(3\right).P\left(-1\right)\ge0\left(Đpcm\right)\)

P/s : Đúng nha 

Q(x) = ax² + bx + c

ta có:Q(-3)=9a+(-3)b+c

Q(1)=a+b+c

Q(-3) - Q(1)=

(9a+(-3)b+c)-(a+b+c)

=(9a-a)+)+(-3b-b)+(c-c)

=8a+(-4)b

= 4.2a+4.-b

=4(2a-b)

thay 2a - b = 0 vào đa thức đã cho, ta được:

Q(-3) - Q(1) =4

=>

Q(-3) - Q(1) >0

mình nhầm 2 dòng cuối nhé phải là

Q(-3) - Q(1) =0

=>

Q(-3) - Q(1) ≥ 0

1 sp cho người trả lời nhanh nhất nha

Ta có:

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

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

\(P\left(3\right)=a.3^2+b.3+c\)

\(\Rightarrow P\left(3\right)=9a+3b+c\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=9a+3b+c-a+b-c\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=8a+4b\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=4\left(2a+b\right)\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=0\)

\(\Rightarrow P\left(3\right)=P\left(-1\right)\)

\(\Rightarrow P\left(-1\right).P\left(3\right)=P\left(3\right)^2\)

Vì \(P\left(3\right)^2\ge0\)

\(\Rightarrow P\left(-1\right).P\left(3\right)\ge0\)

Lời giải:

Ta có:

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

\(\Rightarrow \left\{\begin{matrix} P(-1)=a-b+c\\ P(3)=9a+3b+c\end{matrix}\right.\)

Suy ra: \(P(3)-P(-1)=9a+3b+c-(a-b+c)\)

\(=8a+4b=4(2a+b)=0\)

\(\Rightarrow P(3)=P(-1)\)

\(\Rightarrow P(-1)P(3)=[P(3)]^2\geq 0\)

Ta có đpcm.

2a+b=0=>b=-2a

p(x)=ax^2 -2ax+c

p(-1)=a(-1)^2-2a(-1)+c=3a+c

p(3)=9a-6a+c=3a+c

p(-1).p(3)=(3a+c)^2 >=0=>dpcm

mk thấy đề bài của bn sai rồi 

2a+b=0 ⇒ b=-2a

P(-1)=a(-1)²+(-2a).(-1)+c

        =a+2a+c

        =3a+c

P(3)=a.3²+(-2a).3+c

       =9a-6a+c

       =3a+c

P(-1).P(3)

=(3a+c).(3a+c)

=(3a+c)²

Vì (3a+c)²≥0

⇒P(-1).P(3)≥0

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Lời giải:

Ta có:

$P(1)=a+b+c$

$P(-3)=9a-3b+c$

$P(1)+P(-3)=10a-2b+2c=2(5a-b+c)=2.0=0$

$\Rightarrow P(-3)=-P(1)$

$\Rightarrow P(1)P(-3)=-P(1)^2\leq 0$ 

Ta có đpcm.