Ta có: \(A\left(x\right)=ax^2+bx+c\)
\(\Rightarrow A\left(1\right)=a\cdot1^2+b\cdot1+c=6\Rightarrow a+b+c=6\)
\(a,b,c\) tỉ lệ thuận với \(3,2,1\) suy ra \(\dfrac{a}{3}=\dfrac{b}{2}=\dfrac{c}{1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{3}=\dfrac{b}{2}=\dfrac{c}{1}=\dfrac{a+b+c}{3+2+1}=\dfrac{6}{6}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{3}=1\Rightarrow a=3\\\dfrac{b}{2}=1\Rightarrow b=2\\\dfrac{c}{1}=1\Rightarrow c=1\end{matrix}\right.\)
Ta có : A\(_{\left(x\right)}\)=a2+bx+c
A\(_{\left(1\right)}\)=a+b+c=6
Theo đề bài ta có :
\(\dfrac{a}{3}=\dfrac{b}{2}=c\) (a,b,c\(\ne\)0)
\(\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}=c=\dfrac{a+b+c}{3+2+1}=\dfrac{6}{6}=1\)
Do vậy \(\dfrac{a}{3}=1\Rightarrow a=3\)
\(\dfrac{b}{2}=1\Rightarrow b=2\)
c=1
Vậy a=3, b=2, c=1
