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

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

+ f(5) =a.5² +b.5 +5 =0 => 5a +b =-1 (2)

(10(2) => 4a +(a+b) =-1 => 4a -5 =-1 => 4a =4 => a =1

                                                           => b =-5-a = -5 -1 = -6

Vậy a =1; b =-6 ; c =5

Theo de ta co:

f(0) = a.0²+b.0+c = c =1

f(1)=a.1²+b.1+c = a+b+1 = 2  => a+b = 1

f(2)=a.2²+b.2+c = 4a+2b+1=2(2a+b)+1 = 4  => 2(2a+b) = 3  => 2a+b = 3/2 => b = 3/2 - 2a

Thay b=3/2 - 2a vao bieu thuc: a+b=1  ta duoc:

a+3/2-2a = 1

3/2-a= 1

=> a = 3/2 - 1 = 1/2

Suy ra: b = 3/2 - 2.1/2  = 1/2

Vay: a = 1/2   ;    b=1/2       ;      c=1

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)

\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)

\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)

Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)

f(x)=ax² + bx+ c

f(0)=1, f(1)=2, f(2)=2

=>c=1;a+b+c=2;4a+2b+c=2

=>a+b=1;4a+2b=1

=>a+b=4a+2b

=>4a+2b-a-b=0

=>3a-b=0