Question

[Vu Kim Ngan]

Cho đa thức P(x) = ax² + bx + c và 2a + b = 0. Chứng tỏ rằng P(-1).P(3)\(\ge\)0.

Answer 1

[I don't need you]

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

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

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

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

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

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

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

Answer 2

[Trắng_CV]

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