Ta có:
\(H\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\left\{{}\begin{matrix}H\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c\\H\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}H\left(-1\right)=a-b+c\\H\left(-2\right)=4a-2b+c\end{matrix}\right.\)
\(\Rightarrow H\left(-1\right)+H\left(-2\right)\) \(=\left(a-b+c\right)+\left(4a-2b+c\right)\)
\(=\left(a+4a\right)-\left(b+2b\right)+\left(c+c\right)\)
\(=5a-3b+2c=0\Rightarrow H\left(-1\right)=-H\left(-2\right)\)
\(\Rightarrow H\left(-1\right).H\left(-2\right)=\left[-H\left(-2\right)\right].H\left(-2\right)\)
\(=-H^2\left(-2\right)\)
Mà \(H^2\left(-2\right)\ge0\Leftrightarrow-H^2\left(-2\right)\le0\)
Vậy \(H\left(-1\right).H\left(-2\right)\le0\) (Đpcm)
Nhanh nhá ngày kia nộp!! Lớp viuuuuu~~~