Đáp án: 1
Giải thích các bước giải:
$\begin{array}{l}
a + b + c = 1 \Rightarrow \left\{ \begin{array}{l}
a = 1 - c - b\\
b = 1 - a - c\\
c = 1 - a - b
\end{array} \right.\\
\Rightarrow ab + c = ab + 1 - a - b = a\left( {b - 1} \right) - \left( {b - 1} \right) = \left( {a - 1} \right)\left( {b - 1} \right)\\
= \left( {a - a - b - c} \right)\left( {b - a - b - c} \right)\\
= \left( { - b - c} \right).\left( { - a - c} \right)\\
= \left( {a + c} \right)\left( {b + c} \right)\\
\Rightarrow \left\{ \begin{array}{l}
bc + a = \left( {b + a} \right)\left( {c + a} \right)\\
ac + b = \left( {a + b} \right)\left( {b + c} \right)
\end{array} \right.\\
\Rightarrow P = \frac{{ab + c}}{{{{\left( {a + b} \right)}^2}}}.\frac{{bc + a}}{{{{\left( {b + c} \right)}^2}}}.\frac{{ca + b}}{{{{\left( {c + a} \right)}^2}}}\\
= \frac{{\left( {a + c} \right)\left( {b + c} \right)\left( {a + b} \right)\left( {c + a} \right)\left( {a + b} \right)\left( {b + c} \right)}}{{{{\left( {a + b} \right)}^2}{{\left( {b + c} \right)}^2}{{\left( {a + c} \right)}^2}}}\\
= 1
\end{array}$
