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Lời giải:
Ta có:
$f(-1)=a-b+c$
$f(2)=4a+2b+c$
Cộng lại ta có: $f(-1)+f(2)=5a+b+2c=0$
$\Rightarrow f(-1)=-f(2)$
$\Rightarrow f(-1)f(2)=-f(2)^2\leq 0$ (đpcm)
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)
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: \(f\left(-3\right)\cdot f\left(4\right)=-f\left(4\right)\cdot f\left(4\right)=-\left[f\left(4\right)\right]^2< 0\)
Đề bài bị sai rồi phần đpcm phải là "\(\le\)" chứ không phải "\(< \)
Ta có : \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\hept{\begin{cases}f\left(-3\right)=a.\left(-3\right)^2+b.\left(-3\right)+c=9a-3b+c\\f\left(4\right)=a.4^2+b.4+c=16a+4b+c\end{cases}}\)
\(\Rightarrow f\left(4\right)+f\left(-3\right)=\left(16a+4b+c\right)+\left(9a-3b+c\right)=25a+b+2c=0\)
\(\Rightarrow f\left(-3\right)+f\left(4\right)=0\)
\(\Rightarrow f\left(-3\right)=-f\left(4\right)\)
\(\Rightarrow f\left(-3\right).f\left(4\right)=-f\left(4\right).f\left(4\right)=-[f\left(4\right)]^2\le0\)\(\forall x\)
\(\Rightarrowđpcm\)
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c\)
\(=4a-2b+c\)
\(\Rightarrow f\left(3\right)=a.3^2+b.3+c\)
\(=9a+3b+c\)
\(\Rightarrow f\left(-2\right)+f\left(3\right)=\left(4a-2b+c\right)+\left(9a+3b+c\right)\)
\(=13a+b+2c=0\)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\)
\(\Rightarrow f\left(-2\right).f\left(3\right)\le0\)
Vì \(29a+2c=3b\) => \(c=\frac{3b-29a}{2}\)
Ta có: \(f\left(2\right).f\left(-5\right)=\left[a.2^2+b.2+c\right]\left[a\left(-5\right)^2+b.\left(-5\right)+c\right]\)
\(=\left(4a+2b+c\right)\left(25a-5b+c\right)\)
\(=\left(4a+2b+\frac{3b-29a}{2}\right)\left(25a-5b+\frac{3b-29a}{2}\right)\)
\(=\left(\frac{8a+4b+3b-29a}{2}\right)\left(\frac{50a-10b+3b-29a}{2}\right)\)
\(=\left(\frac{-21a+7b}{2}\right)\left(\frac{21a-7b}{2}\right)\)
\(=\frac{-7}{2}\left(3a-b\right).\frac{7}{2}\left(3a-b\right)\)
\(=\frac{-49}{4}\left(3a-b\right)^2\le0\) (ĐFCM)
Lời giải:
Ta có:
$f(4)=16a+4b+c$
$f(-2)=4a-2b+c$
Cộng theo vế: $f(4)+f(-2)=20a+2b+2c=2(10a+b+c)=2.0=0$
$\Rightarrow f(-2)=-f(4)$
$\Rightarrow f(4).f(-2)=f(4).-f(4)=-f(4)^2\leq 0$
Ta có đpcm.