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đối xứng qua x=0
=> b=0
4a+c=0
c=-4a
c<a=> -4a<a => a>0
chưa thể xác định chính xác đươc
đa thức là ax^2-4a=0 với a>0
Ta có:
\(f\left(0\right)=a.0^2+b.0+c=0\)
\(=0+0+c=0\Rightarrow c=0\)
\(f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=0\)
\(a-b+0=0\)
\(\Rightarrow a-b=0\)
\(\Rightarrow a=b\)
\(f\left(1\right)=a.1^2+b.1+c=0\)
\(\Rightarrow a+b+0=0\)
\(\Rightarrow a+b=0\)
Mà \(a=b\)
\(\Rightarrow a=b=\frac{0}{2}=0\)
Vậy \(a=b=c=0\)
\(f\left(0\right)=2010\Rightarrow a.0^2+b.0+c=2010\Rightarrow c=2010\)
\(f\left(1\right)=2011\Rightarrow a.1^2+b.1+c=2011\Rightarrow a+b+c=2011\)
\(\Rightarrow a+b+2010=2011\Rightarrow a+b=1\) (1)
\(f\left(-1\right)=2012\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=2012\)
\(\Rightarrow a-b+c=2012\Rightarrow a-b+2010=2012\)
\(\Rightarrow a-b=2\Rightarrow a=b+2\)
Thế vào (1) \(\Rightarrow b+2+b=1\Rightarrow2b=-1\Rightarrow b=-\dfrac{1}{2}\)
\(\Rightarrow a=b+2=-\dfrac{1}{2}+2=\dfrac{3}{2}\)
\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^2-\dfrac{1}{2}x+2010\)
\(\Rightarrow f\left(-2\right)=\dfrac{3}{2}.\left(-2\right)^2-\dfrac{1}{2}.\left(-2\right)+2010=2017\)
\(\left\{{}\begin{matrix}f\left(0\right)=5\Rightarrow0+0+5\Rightarrow c=5\\f\left(1\right)=0\Rightarrow a+b+5=0\\f\left(5\right)=0\Rightarrow25a+5b+5=0\end{matrix}\right.\) \(\left\{{}\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\end{matrix}\right.\)
tu (3) => b =-1-5a
tu (2) => a-1-5a+5 =0 => a =1 ;b =-6
y =x^2 -6x +5
y(-1) =1 +6 +5 khac 3 => loai
y(-1/2) =1/4 -6/2 +5 =1/4 +2 = 9/4 nhan
Q(1/2;9/4) thuoc dths
Ta có: \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(0\right)=c⋮3\Rightarrow c⋮3\)
\(\left\{{}\begin{matrix}f\left(1\right)=a+b+c⋮3\\f\left(-1\right)=a-b+c⋮3\end{matrix}\right.\)
Mà \(c⋮5\)
\(\Rightarrow\left\{{}\begin{matrix}a+b⋮3\\a-b⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a⋮3\\2b⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a⋮3\\b⋮3\end{matrix}\right.\) ( do \(\left(2;3\right)=1\) )
Vậy \(a,b,c⋮3\)
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)
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ta có f(2)=0 =>4a2+2b+c=0 => 4a2+2b=-c (1)
f(-2)=0 => 4a2- 2b+c=0 => 4a2-2b=-c (2)
từ (1), (2) => a=0, b=1, c=-2