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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(−4)=16a−4b+c
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⇒f(−4)−6f(−1)=16a−4b+c−6(a−b+c)=10a+2b−5c=0
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⇒f(−4)=6f(−1)
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⇒f(−1)f(−4)=f(−1).6f(−1)=6[f(−1)]
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≥0 (đpcm)
b.
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f(−2)=4a−2b+c
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f(3)=9a+3b+c
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⇒f(−2)+f(3)=13a+b+2c=0
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⇒f(−2)=−f(3)
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⇒f(−2)f(3)=−[f(3)]
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≤0 (đpcm
\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)
\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)
Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)
f(0) = 1
\(\Rightarrow\) a.02 + b.0 + c = 1
\(\Rightarrow\) c = 1
Vậy hệ số a = 0; b = 0; c = 1
f(1) = 2
\(\Rightarrow\) a.12 + b.1 + c = 2
\(\Rightarrow\) a + b + c = 2
Vậy hệ số a = 1; b = 1; c = 1
f(2) = 4
\(\Rightarrow\) a.22 + b.2 + c = 4
\(\Rightarrow\) 4a + 2b + c = 4
Vậy hệ số a = 4; b = 2; c = 1
Chúc bn học tốt! (chắc vậy :D)
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{4}a+\frac{1}{2}b+c\)
\(\Rightarrow f\left(-2\right)=4a-2b+c\)
\(\Rightarrow f\left(\frac{1}{2}\right)+f\left(-2\right)=\frac{17}{4}a-\frac{3}{2}b+2c\)
\(\Rightarrow4\left[f\left(\frac{1}{2}\right)+f\left(-2\right)\right]=17a-6b+8c=0\)( vì 17a-6b+8c=0)
\(\Rightarrow f\left(\frac{1}{2}\right)+f\left(-2\right)=0\)
\(\Rightarrow f\left(\frac{1}{2}\right)=-f\left(-2\right)\)
\(\Rightarrow f\left(\frac{1}{2}\right).f\left(-2\right)=-\left[f\left(-2\right)\right]^2\le0\left(đpcm\right)\)
\(f\left(2\right)=a.2^2+b.2+c=4a+2b+c=10a-10b-\left(6a-12b-c\right)=10a-10b\)
\(f\left(-3\right)=a.\left(-3\right)^2+b.\left(-3\right)+c=9a-3b+c=15a-15b-\left(6a-12b-c\right)=15a-15b\)
\(\Rightarrow f\left(2\right).f\left(-3\right)=\left(10a-10b\right).\left(15a-15b\right)=150\left(a-b\right)^2\)
Mà \(\left(a-b\right)^2\ge0;\forall a;b\Rightarrow150\left(a-b\right)^2\ge0\)
\(\Rightarrow f\left(2\right).f\left(-3\right)\ge0\)
\(f\left(-1\right)=a\left(-1\right)^2+b.\left(-1\right)+c\)
\(=a-b+c\)
\(f\left(2\right)=a.2^2+b.2+c\)
\(=4a+2b+c\)
\(\Rightarrow f\left(2\right)-2.f\left(-1\right)=\left(4a+2b+c\right)-2\left(a-b+c\right)\)
\(=2a+4b-c=0\)
\(\Rightarrow f\left(2\right)=2.f\left(-1\right)\)
\(\Rightarrow f\left(2\right)\)và \(2.f\left(-1\right)\)cùng dấu
\(\Rightarrow f\left(2\right)\)và \(f\left(-1\right)\)cùng dấu
\(\Rightarrow f\left(2\right).f\left(-1\right)\ge0\)(đpcm)
Ta có :\(f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c\)
\(f\left(2\right)=a.2^2+b.2+c=4a+2b+c\)
\(\implies\) \(f\left(2\right)-2f\left(-1\right)=\left(4a+2b+c\right)-2.\left(a-b+c\right)\)
\(\implies\) \(f\left(2\right)=2.f\left(-1\right)\)
\(\implies\) \(f\left(-1\right).f\left(2\right)=f\left(-1\right).2f\left(-1\right)=f\left(-1\right)^2.2\) \(\geq\) \(0\)
\(\implies\) \(f\left(-1\right).f\left(2\right)\) \(\geq\) \(0\) \(\left(đpcm\right)\)