Đáp án:
$\begin{array}{l}
Do:ab + bc + ca = 1\\
\Rightarrow 1 + {c^2} = ab + bc + ca + {c^2}\\
= b\left( {a + c} \right) + c\left( {a + c} \right)\\
= \left( {a + c} \right)\left( {b + c} \right)\\
1 + {a^2} = ab + bc + ca + {a^2}\\
= \left( {a + c} \right)\left( {b + a} \right)\\
1 + {b^2} = ab + bc + ca + {b^2}\\
= c\left( {b + a} \right) + b\left( {b + a} \right)\\
= \left( {b + a} \right)\left( {c + b} \right)\\
\Rightarrow \dfrac{{a - b}}{{1 + {c^2}}} + \dfrac{{b - c}}{{1 + {a^2}}} + \dfrac{{c - a}}{{1 + {b^2}}}\\
= \dfrac{{a - b}}{{\left( {c + a} \right)\left( {c + b} \right)}} + \dfrac{{b - c}}{{\left( {a + b} \right)\left( {a + c} \right)}}\\
+ \dfrac{{c - a}}{{\left( {b + a} \right)\left( {b + c} \right)}}\\
= \dfrac{{\left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {a + c} \right)}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}}\\
= \dfrac{{{a^2} - {b^2} + {b^2} - {c^2} + {c^2} - {a^2}}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}}\\
= 0
\end{array}$
