Ta có: \(f\left(0\right)=a.0^2+b.0+c=1\Rightarrow c=1.\)
Lại có: \(f\left(1\right)=a.1^2+b.1+1=2\Rightarrow a+b=1.\)
\(f\left(2\right)=2^2.a+2b+1=2\)
\(\Rightarrow4a+2b=a+b\)
\(\Rightarrow3a+b=0\)\(\Rightarrow6a+2b=0\)\(\Rightarrow2a=-1\Rightarrow a=-0,5\)
\(\Rightarrow b=1,5\)
Vậy: \(a=-0,5\)
\(b=1,5\)
\(c=1\)
\(\)
thay f(0) = 1 vào đa thức ta có : \(a.0^2+b0+c=1\Leftrightarrow c=1\)
tiếp tục thay f(1) = 2 và f(2) = 2 vào đa thức
ta có : hệ phương trình \(\left\{{}\begin{matrix}a+b+c=2\\4a+2b+c=2\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b+1=2\\4a+2b+1=2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=1\\4a+2b=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2a+2b=2\\4a+2b=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}2a=-1\\a+b=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=\dfrac{-1}{2}\\\dfrac{-1}{2}+b=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=\dfrac{-1}{2}\\b=\dfrac{3}{2}\end{matrix}\right.\)
vậy \(a=\dfrac{-1}{2};b=\dfrac{3}{2};c=1\)
