Question

cho ba so a b c thoa man 25a+b+2c=0. Dat f(x)=\(ax^2+bx+c\)chung minh f(-3).f(4)<o

 

Answer 1

Ta có: \(f\left(x\right)=ax^2+bx+c\)

\(\Rightarrow\hept{\begin{cases}f\left(-3\right)=9a-3b+c\\f\left(4\right)=16a+4a+c\end{cases}}\) \(\Rightarrow f\left(-3\right)+f\left(4\right)=25a+b+2c=0\)

\(\Rightarrow f\left(-3\right)=-f\left(4\right)\)

Khi đó: \(f\left(-3\right)\cdot f\left(4\right)=-f\left(4\right)\cdot f\left(4\right)=-\left[f\left(4\right)\right]^2< 0\)

Answer 2

Đề bài bị sai rồi phần đpcm phải là "\(\le\)" chứ không phải "\(< \)

Ta có : \(f\left(x\right)=ax^2+bx+c\)

\(\Rightarrow\hept{\begin{cases}f\left(-3\right)=a.\left(-3\right)^2+b.\left(-3\right)+c=9a-3b+c\\f\left(4\right)=a.4^2+b.4+c=16a+4b+c\end{cases}}\)

\(\Rightarrow f\left(4\right)+f\left(-3\right)=\left(16a+4b+c\right)+\left(9a-3b+c\right)=25a+b+2c=0\)

\(\Rightarrow f\left(-3\right)+f\left(4\right)=0\)

\(\Rightarrow f\left(-3\right)=-f\left(4\right)\)

\(\Rightarrow f\left(-3\right).f\left(4\right)=-f\left(4\right).f\left(4\right)=-[f\left(4\right)]^2\le0\)\(\forall x\)

\(\Rightarrowđpcm\)