Đáp án:
Quy đồng mẫu số của vế trái ta được
$\begin{array}{l}
\frac{{x - y}}{{1 + xy}} + \frac{{y - z}}{{1 + yz}} + \frac{{z - x}}{{1 + zx}}\\
= \frac{{\left( {x - y} \right)\left( {1 + yz} \right)\left( {1 + zx} \right) + \left( {y - z} \right)\left( {1 + xy} \right)\left( {1 + zx} \right) + \left( {z - x} \right)\left( {1 + xy} \right)\left( {1 + yz} \right)}}{{\left( {1 + xy} \right)\left( {1 + yz} \right)\left( {1 + zx} \right)}}\\
TS = \left( {x - y} \right)\left( {1 + yz} \right)\left( {1 + zx} \right)\\
+ \left( {1 + xy} \right)\left[ {\left( {y - z} \right)\left( {1 + zx} \right) + \left( {z - x} \right)\left( {1 + yz} \right)} \right]\\
= \left( {x - y} \right)\left( {1 + yz} \right)\left( {1 + zx} \right)\\
+ \left( {1 + xy} \right)\left[ {y - z + xyz - x{z^2} + z - x + y{z^2} - xyz} \right]\\
= \left( {x - y} \right)\left( {1 + yz} \right)\left( {1 + zx} \right)\\
+ \left( {1 + xy} \right)\left[ {y - x{z^2} - x + y{z^2}} \right]\\
= \left( {x - y} \right)\left( {1 + yz} \right)\left( {1 + zx} \right) + \left( {1 + xy} \right)\left( {x - y} \right)\left( { - 1 - {z^2}} \right)\\
= \left( {x - y} \right)\left[ {\left( {1 + yz} \right)\left( {1 + zx} \right) + \left( {1 + xy} \right)\left( { - 1 - {z^2}} \right)} \right]\\
= \left( {x - y} \right)\left( {1 + xz + yz + xy{z^2} - 1 - {z^2} - xy - xy{z^2}} \right)\\
= \left( {x - y} \right)\left( {xz + yz - {z^2} - xy} \right)\\
= \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\\
Vậy\,\frac{{x - y}}{{1 + xy}} + \frac{{y - z}}{{1 + yz}} + \frac{{z - x}}{{1 + zx}} = \frac{{x - y}}{{1 + xy}}.\frac{{y - z}}{{1 + yz}}.\frac{{z - x}}{{1 + zx}}
\end{array}$
