Đáp án:
\[A = \frac{3}{2}\]
Giải thích các bước giải:
Ta có:
\[a + b + c = 0 \Leftrightarrow \left\{ \begin{array}{l}
a = - \left( {b + c} \right)\\
b = - \left( {c + a} \right)\\
c = - \left( {b + a} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{a^2} = {b^2} + 2bc + {c^2}\\
{b^2} = {c^2} + 2ca + {a^2}\\
{c^2} = {a^2} + 2ab + {b^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{a^2} - {b^2} - {c^2} = 2bc\\
{b^2} - {c^2} - {a^2} = 2ca\\
{c^2} - {a^2} - {b^2} = 2ab
\end{array} \right.\]
Lại có:
\(\begin{array}{l}
{a^3} + {b^3} + {c^3} = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc + ca} \right) + 3abc\\
\Rightarrow {a^3} + {b^3} + {c^3} = 3abc
\end{array}\)
Suy ra:
\(\begin{array}{l}
A = \frac{{{a^2}}}{{{a^2} - {b^2} - {c^2}}} + \frac{{{b^2}}}{{{b^2} - {c^2} - {a^2}}} + \frac{{{c^2}}}{{{c^2} - {a^2} - {b^2}}}\\
= \frac{{{a^2}}}{{2bc}} + \frac{{{b^2}}}{{2ca}} + \frac{{{c^2}}}{{2ab}}\\
= \frac{{{a^3} + {b^3} + {c^3}}}{{2abc}} = \frac{{3abc}}{{2abc}} = \frac{3}{2}
\end{array}\)
