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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)
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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b.
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⇒f(−2)f(3)=−[f(3)]
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Ta có: f(0) = a.02 + b.0 + c = 2
=> c = 2
f(1) = a.12 + b.1 + c = 1
=> a + b + c = 1 => a + b = 1 - c = 1 - 2 = -1 (1)
f(-2) = a.(-2)2 + b.(-2) + c = 2
=> 4a - 2b = 2 - c = 2 - 2 = 0
=> 2a - b = 0 (2)
Từ (1) và (2) cộng vế theo vế:
(a + b) + (2a - b) = -1
=> 3a = -1
=> a = -1/3
=> b = -1 - a = -1 + 1/3 = -2/3
Vậy ....
\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)
\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)
Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)
Vì f(0)=5 nên x*0+b*0+c=5
0+0+c=5 nên c=5
Vì f(1)=0 nên a*12+b*1+5=0
a+b+5=0
a+b=0-5
a+b=-5
Vì f(5)=0 nên a*52+b*5+5=0
5(5a+b+1)=0
5a+b+1=0/5=0
4a+a+b=0-1
4a+(-5)=-1
4a=-1-(-5)
4a=4
a=4/4
a=1
nên b=-5-1=-6
Vậy a=1;b=-6 và c=5
Ta co:
- f(0) = a.02+b.0+c = 0+0+c = c= 5
- f(1) = a.12+b.1+c = a+b+5 = 0 => a+b = -5
- f(5) = a.52+b.5+c = 25a + 5b + 5 = 0 => 25a+5b = -5
=> a+b = 25a+5b = -5
=> 25a-a + 5b-b = 0
=> 24a + 4b = 0
=> 24a = -4b
=> 24/-4 = b/a
=> b/a = -6
Tu \(\frac{b}{a}=-6=>\frac{b}{-6}=\frac{a}{1}=\frac{b+a}{-6+1}=-\frac{5}{-5}=1\)
=> a = 1 ; b=-6
Vay: a=1 ; b=-6 ; c =5
\(f\left(x_1\right)=ax_1\) ; \(f\left(x_2\right)=ax_2\) ; \(f\left(x_1x_2\right)=ax_1x_2\)
Để \(f\left(x_1\right)f\left(x_2\right)=f\left(x_1x_2\right)\)
\(\Leftrightarrow ax_1.ax_2=ax_1x_2\)
\(\Leftrightarrow a^2x_1x_2=ax_1x_2\)
\(\Leftrightarrow a^2=a\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=1\end{matrix}\right.\)
Vậy \(a=1\)
\(f\left(x\right)=ax^2+bx+c\)
\(f\left(2\right)=4a+2b+c\)
\(f\left(-1\right)=a-b+c\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=4a+2b+c+a-b+c\)
\(\Leftrightarrow f\left(2\right)+f\left(-1\right)=5a+b+2c=0\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=0\Leftrightarrow f\left(2\right)=-f\left(-1\right)\)
\(\Leftrightarrow f\left(2\right).f\left(-1\right)=-f\left(-1\right).f\left(-1\right)\le0\)
\(\Rightarrowđpcm\)
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(x-1\right)=a\left(x-1\right)^2+b\left(x-1\right)+c\)
\(\Rightarrow f\left(x\right)-f\left(x-1\right)=ax^2+bx+c-ax^2+2ax-a-bx+b-c=x\)
\(\Leftrightarrow2ax-a+b-x=0\)
\(\Leftrightarrow\left(2a-1\right)x+b-a=0\)
\(\Leftrightarrow\hept{\begin{cases}2a-1=0\\b-a=0\end{cases}\Leftrightarrow}a=b=\frac{1}{2}\)
\(\)và Hàm số đúng với mọi giá trị của \(c\)
Vậy \(a=b=\frac{1}{2};c\in R\)