a) Giải:
Ta có:
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c\\f\left(3\right)=a.3^2+b.3+c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(-2\right)=4a-2b+c\\f\left(3\right)=9a+3b+c\end{matrix}\right.\)
\(\Rightarrow f\left(-2\right)+f\left(3\right)=\left(4a-2b+c\right)+\left(9a+3b+c\right)\)
\(=\left(4a+9a\right)+\left(-2b+3b\right)+\left(c+c\right)\)
\(=13a+b+2c=0\)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\)
\(\Rightarrow f\left(-2\right).f\left(3\right)=-\left[f\left(3\right)\right]^2\le0\)
Vậy \(f\left(-2\right).f\left(3\right)\le0\) (Đpcm)
b) Sửa đề:
Biết \(5a+b+2c=0\)
Giải:
Ta có:
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=a.2^2+b.2+c=4a+2b+c\\f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c\end{matrix}\right.\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=\left(a-b+c\right)+\left(4a+2b+c\right)\)
\(=\left(4a+a\right)+\left(-b+2b\right)+\left(c+c\right)\)
\(=5a+b+2c=0\)
\(\Rightarrow f\left(2\right)=-f\left(-1\right)\)
\(\Rightarrow f\left(2\right).f\left(-1\right)=-\left[f\left(-1\right)\right]^2\le0\)
Vậy \(f\left(2\right).f\left(-1\right)\le0\) (Đpcm)
