Đáp án:
$\begin{array}{l}
A,B \in \left( P \right)\\
\Rightarrow \left\{ \begin{array}{l}
5 = a.0 + b.0 + c\\
2 = a.{\left( { - 1} \right)^2} + b.\left( { - 1} \right) + c
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
c = 5\\
a - b = - 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
c = 5\\
a = b - 3
\end{array} \right.\\
Dinh\,I\left( {\dfrac{{ - b}}{{2a}};0} \right)\left( {do:I \in Ox} \right)\\
\Rightarrow 0 = a.{\left( {\dfrac{{ - b}}{{2a}}} \right)^2} + b.\left( {\dfrac{{ - b}}{{2a}}} \right) + 5\\
\Rightarrow a.\dfrac{{{b^2}}}{{4{a^2}}} - \dfrac{{{b^2}}}{{2a}} + 5 = 0\\
\Rightarrow \dfrac{{{b^2}}}{{4a}} - \dfrac{{{b^2}}}{{2a}} = - 5\\
\Rightarrow \dfrac{{ - {b^2}}}{{4a}} = - 5\\
\Rightarrow {b^2} = 20a\\
Do:a = b - 3\\
\Rightarrow {b^2} = 20.\left( {b - 3} \right)\\
\Rightarrow {b^2} - 20b + 60 = 0\\
\Rightarrow \left[ \begin{array}{l}
b = 10 + 2\sqrt {10} \Rightarrow a = 7 + 2\sqrt {10} \\
b = 10 - 2\sqrt {10} \Rightarrow a = 7 - 2\sqrt {10}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
a + b + c = 22 + 4\sqrt {10} \\
a + b + c = 22 - 4\sqrt {10}
\end{array} \right.
\end{array}$
