Đáp án:
\[a = - 3;\,\,\,\,b = 7;\,\,\,c = - 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P\left( x \right) = a{x^2} + bx + c\\
P\left( 0 \right) = - 1 \Leftrightarrow a{.0^2} + b.0 + c = - 1 \Leftrightarrow c = - 1\\
P\left( 1 \right) = 3 \Leftrightarrow a{.1^2} + b.1 + c = 3 \Leftrightarrow a + b + c = 3\\
P\left( 2 \right) = 1 \Leftrightarrow a{.2^2} + b.2 + c = 1 \Leftrightarrow 4a + 2b + c = 1\\
\Rightarrow \left\{ \begin{array}{l}
c = - 1\\
a + b + c = 3\\
4a + 2b + c = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
c = - 1\\
a + b + \left( { - 1} \right) = 3\\
4a + 2b + \left( { - 1} \right) = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
c = - 1\\
a + b = 4\\
4a + 2b = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
c = - 1\\
a + b = 4\\
2a + b = 1
\end{array} \right.\\
\Rightarrow \left( {2a + b} \right) - \left( {a + b} \right) = 1 - 4\\
\Leftrightarrow a = - 3\\
\Rightarrow b = 4 - a = 7\\
\Rightarrow a = - 3;\,\,\,\,b = 7;\,\,\,c = - 1
\end{array}\)
Vậy \(a = - 3;\,\,\,\,b = 7;\,\,\,c = - 1\)
