Question

Cho đa thức: \(f\left(x\right)=ax^2+bx+c\).

Chứng tỏ: \(f\left(-2\right)\times f\left(3\right)\le0\)

biết \(13a+b+2c=0\)

Giúp mk nha!!!Cảm ơn nhiều!!!

Answer 1

Ta có \(f\left(-2\right)\times f\left(-3\right)=\left(4a-2b+c\right).\left(9a+3b+c\right)=\left(4a-2b+c\right).\left[13a+b+2c-\left(4a-2b+c\right)\right]\)

Mà \(13a+b+2c=0\) theo giả thiết.

\(\Rightarrow f\left(-2\right)\times f\left(3\right)=-\left[\left(4a-2b+c\right)^2\right]\)

\(\left(4a-2b+c\right)^2\) luôn \(\ge0\Rightarrow f\left(-2\right)\times f\left(3\right)\) \(\le0\)