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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)
\(f\left(2\right)=a.2^2+b.2+c=4a+2b+c\)
\(f\left(-5\right)=a.\left(-5\right)^2+b.\left(-5\right)+c=25a-5b+c\)
\(f\left(2\right)+f\left(5\right)=4a+2b+c+25a-5b+c=29a-3b+2c\)
\(=\left(29a+2c\right)-3b=3b-3b=0\)
\(\Leftrightarrow f\left(2\right)=-f\left(-5\right)\)
\(\Leftrightarrow f\left(2\right)f\left(-5\right)\le0\).
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(−1)f(−4)=f(−1).6f(−1)=6[f(−1)]
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⇒f(−2)f(3)=−[f(3)]
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\(f\left(x\right)=ax^2+bx+c\)
\(f\left(2\right)=4a+2b+c\)
\(f\left(-1\right)=a-b+c\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=4a+2b+c+a-b+c\)
\(\Leftrightarrow f\left(2\right)+f\left(-1\right)=5a+b+2c=0\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=0\Leftrightarrow f\left(2\right)=-f\left(-1\right)\)
\(\Leftrightarrow f\left(2\right).f\left(-1\right)=-f\left(-1\right).f\left(-1\right)\le0\)
\(\Rightarrowđpcm\)
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.
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)
\(f\left(-1\right)=a+c-b\)
\(f\left(3\right)=9a+3b+c=10a+2b+2c+b-a-c=b-a-c\)
\(\Rightarrow f\left(-1\right).f\left(3\right)=\left(a+c-b\right)\left(b-a-c\right)=-\left(a+c-b\right)^2\le0\)
\(f\left(2\right)=a.2^2+b.2+c=4a+2b+c=10a-10b-\left(6a-12b-c\right)=10a-10b\)
\(f\left(-3\right)=a.\left(-3\right)^2+b.\left(-3\right)+c=9a-3b+c=15a-15b-\left(6a-12b-c\right)=15a-15b\)
\(\Rightarrow f\left(2\right).f\left(-3\right)=\left(10a-10b\right).\left(15a-15b\right)=150\left(a-b\right)^2\)
Mà \(\left(a-b\right)^2\ge0;\forall a;b\Rightarrow150\left(a-b\right)^2\ge0\)
\(\Rightarrow f\left(2\right).f\left(-3\right)\ge0\)