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)

a) \(f\left(0\right)=a.0+b=-3\Rightarrow b=-3\)

\(f\left(1\right)=a.1+b=a+b=-3\Rightarrow a=2\)

b) \(f\left(x\right)=2x-3\)

f(x1+x2)= a(x1+x2)+b=ax1+ax2+b. Tương tự ta có f(x1)+f(x2)=ax1+b+ax2+b. => để 2 vế bằng nhau thì b=0

đúng ko bạn

:))))

:)))))))))))))))

\(f\left(0\right)=5=>c=5;f\left(2\right)=4.a+2.b+5=0;f\left(5\right)=25a+5b+5=0\Leftrightarrow5a+b+1=0\)

\(\hept{\begin{cases}4a+2b+5=0\\5a+b+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4a+2b+5=0\\10a+2b+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4a+2b+5=0\\6a-3=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}b=-\frac{7}{2}\\a=\frac{1}{2}\end{cases}}\)

\(f\left(x\right)=\frac{1}{2}x^2-\frac{7}{2}x+5\)

b)

\(f\left(-1\right)=\frac{1}{2}+\frac{7}{2}+5=9=>P\left(-1;3\right)kothuocHS\)

\(f\left(\frac{1}{2}\right)=\frac{1}{2}.\frac{1}{4}-\frac{7}{2}.\frac{1}{2}+5=\frac{\left(1-14+5.8\right)}{8}=\frac{27}{8}=>Qkothuoc\)

c)

\(\frac{1}{2}x^2-\frac{7}{2}x+5=-3\Rightarrow\frac{1}{2}x^2-\frac{7}{2}x+8=0\)

\(x^2-7x+16=0\Leftrightarrow\left(x^2-2.\frac{7}{2}x+\frac{49}{4}\right)+\frac{15}{4}\)vo nghiem