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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)
Lời giải:
a.
$f(-1)=a-b+c$
$f(-4)=16a-4b+c$
$\Rightarrow f(-4)-6f(-1)=16a-4b+c-6(a-b+c)=10a+2b-5c=0$
$\Rightarrow f(-4)=6f(-1)$
$\Rightarrow f(-1)f(-4)=f(-1).6f(-1)=6[f(-1)]^2\geq 0$ (đpcm)
b.
$f(-2)=4a-2b+c$
$f(3)=9a+3b+c$
$\Rightarrow f(-2)+f(3)=13a+b+2c=0$
$\Rightarrow f(-2)=-f(3)$
$\Rightarrow f(-2)f(3)=-[f(3)]^2\leq 0$ (đpcm)
a.
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⇒f(−2)f(3)=−[f(3)]
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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.
\(f\left(2\right)=a.2^2+b.2+c=4a+2b+c=10a-10b-\left(6a-12b-c\right)=10a-10b\)
\(f\left(-3\right)=a.\left(-3\right)^2+b.\left(-3\right)+c=9a-3b+c=15a-15b-\left(6a-12b-c\right)=15a-15b\)
\(\Rightarrow f\left(2\right).f\left(-3\right)=\left(10a-10b\right).\left(15a-15b\right)=150\left(a-b\right)^2\)
Mà \(\left(a-b\right)^2\ge0;\forall a;b\Rightarrow150\left(a-b\right)^2\ge0\)
\(\Rightarrow f\left(2\right).f\left(-3\right)\ge0\)
\(f\left(x\right)=ax^2+bx+c\)
\(f\left(2\right)=4a+2b+c\)
\(f\left(-1\right)=a-b+c\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=4a+2b+c+a-b+c\)
\(\Leftrightarrow f\left(2\right)+f\left(-1\right)=5a+b+2c=0\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=0\Leftrightarrow f\left(2\right)=-f\left(-1\right)\)
\(\Leftrightarrow f\left(2\right).f\left(-1\right)=-f\left(-1\right).f\left(-1\right)\le0\)
\(\Rightarrowđpcm\)
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 \(\ge0\Rightarrow f\left(-2\right)\times f\left(3\right)\) \(\le0\)
\(f\left(-1\right)=a+c-b\)
\(f\left(3\right)=9a+3b+c=10a+2b+2c+b-a-c=b-a-c\)
\(\Rightarrow f\left(-1\right).f\left(3\right)=\left(a+c-b\right)\left(b-a-c\right)=-\left(a+c-b\right)^2\le0\)