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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)
\(f\left(x\right)=ax^2+bx+c\)
\(f\left(2\right)=4a+2b+c\)
\(f\left(-1\right)=a-b+c\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=4a+2b+c+a-b+c\)
\(\Leftrightarrow f\left(2\right)+f\left(-1\right)=5a+b+2c=0\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=0\Leftrightarrow f\left(2\right)=-f\left(-1\right)\)
\(\Leftrightarrow f\left(2\right).f\left(-1\right)=-f\left(-1\right).f\left(-1\right)\le0\)
\(\Rightarrowđpcm\)
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 \(\ge0\Rightarrow f\left(-2\right)\times f\left(3\right)\) \(\le0\)
Ta có \(f\left(-2\right).f\left(3\right)=\left(4a-2b+c\right)\left(9a+3b+c\right)\)
\(=36a^2-6b^2+c^2-6ab+13ac+bc\)
Thay b = - 13a - 2c, ta có
\(36a^2-6\left(-13a-2c\right)^2+c^2-6a\left(-13a-2c\right)+13ac+\left(-13a-2c\right)c\)
\(=-900a^2-300ac-25c^2=-25\left(36a^2+12ac+c^2\right)\)
\(-25\left(6a+c\right)^2\le0\forall a;c\)
Vậy nên \(f\left(-2\right).f\left(3\right)\le0\)
Cách này đơn giản hơn: Có \(f\left(-2\right)=4a-2b+c;f\left(3\right)=9a+3b+c\)
Do đó \(f\left(-2\right)+f\left(3\right)=13a+b+2c=0\) (theo giả thiết). Từ đó \(f\left(-2\right)=-f\left(3\right)\) nên
\(f\left(-2\right)f\left(3\right)=-f^2\left(3\right)\le0\)
Ta có:
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(2\right).f\left(-3\right)=\left(4a-2b+c\right)\left(9a+3b+c\right)\)
\(=\left(4a-2b+c\right)\left[13a+b+2c-\left(4a-2b+c\right)\right]\)
Mà \(13a+b+c=0\)
\(\Rightarrow f\left(2\right).f\left(-3\right)=-\left[\left(4a-2b+c\right)^2\right]\)
Ta có \(\left(4a-2b+c\right)^2\ge0\Rightarrow-\left[\left(4a-2b+c\right)^2\right]\le0\)
Vậy nếu \(13a+b+2c=0\)\(\Rightarrow f\left(2\right).f\left(3\right)\le0\) (Đpcm)
\(f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=4a-2b+c\)
\(f\left(3\right)=a.3^2+b.3+c=9a+3b+c\)
\(f\left(-2\right)+f\left(3\right)=13a+b+2c=0\)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\Rightarrow f\left(-2\right).f\left(3\right)\le0\)