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(0)=a0+b0+c=2010

=>c=2010

f(1)=a1+b1+c=a1+b1+2010

=>a+b=1 (1)

f(-1)=a1+(-b1)+c=a1-b1+2010

=>a-b=2 (2)

Từ (1) và (2) => a=(2+1):2=1,5

                        b=(1-2):2=-0,5

Vậy f(2)=1,5.2+(-0,5)x2+2010=2014

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 ....

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. 

�
(
−
1
)
=
�
−
�
+
�
f(−1)=a−b+c

�
(
−
4
)
=
16
�
−
4
�
+
�
f(−4)=16a−4b+c

⇒
�
(
−
4
)
−
6
�
(
−
1
)
=
16
�
−
4
�
+
�
−
6
(
�
−
�
+
�
)
=
10
�
+
2
�
−
5
�
=
0
⇒f(−4)−6f(−1)=16a−4b+c−6(a−b+c)=10a+2b−5c=0

⇒
�
(
−
4
)
=
6
�
(
−
1
)
⇒f(−4)=6f(−1)

⇒
�
(
−
1
)
�
(
−
4
)
=
�
(
−
1
)
.
6
�
(
−
1
)
=
6
[
�
(
−
1
)
]
2
≥
0
⇒f(−1)f(−4)=f(−1).6f(−1)=6[f(−1)] 
2
 ≥0 (đpcm)

b.

�
(
−
2
)
=
4
�
−
2
�
+
�
f(−2)=4a−2b+c

�
(
3
)
=
9
�
+
3
�
+
�
f(3)=9a+3b+c

⇒
�
(
−
2
)
+
�
(
3
)
=
13
�
+
�
+
2
�
=
0
⇒f(−2)+f(3)=13a+b+2c=0

⇒
�
(
−
2
)
=
−
�
(
3
)
⇒f(−2)=−f(3)

⇒
�
(
−
2
)
�
(
3
)
=
−
[
�
(
3
)
]
2
≤
0
⇒f(−2)f(3)=−[f(3)] 
2
 ≤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 \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)

\(f\left(2\right)=a.2^2+b.2+c=4a+2b+c=10a-10b-\left(6a-12b-c\right)=10a-10b\)

\(f\left(-3\right)=a.\left(-3\right)^2+b.\left(-3\right)+c=9a-3b+c=15a-15b-\left(6a-12b-c\right)=15a-15b\)

\(\Rightarrow f\left(2\right).f\left(-3\right)=\left(10a-10b\right).\left(15a-15b\right)=150\left(a-b\right)^2\)

Mà \(\left(a-b\right)^2\ge0;\forall a;b\Rightarrow150\left(a-b\right)^2\ge0\)

\(\Rightarrow f\left(2\right).f\left(-3\right)\ge0\)

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: 

=> 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

Vơí f (0) = 5 => 5 = a.0² +b.0 + c => 0+ c =5 => c = 5

Với f (1) =0 => 0 = a. 1²+ b.1+ 5 => a +b = - 5

Với f(5) =0 => 0 = a.5² + b.5 + 5 => (5a + b + 1) .5 = 0

                                                =>. 5a + b + 1= 0

                                                => 4a + a +b = -1

                                                => 4a + (-5) = -1

                                                => 4a = 4  => a = 1

                                                => b = - 6

Vậy a = 1; b = - 6; c = 5

Tik đúng nha