Nếu \(\frac{1}{{b + c}};\,\,\frac{1}{{c + a}};\,\...

D

- Hướng dẫn giải

$\begin{array}{l} \frac{2}{{c + a}} = \frac{1}{{b + c}} = \frac{1}{{a + b}} \Leftrightarrow \frac{{c + a}}{2} = \frac{{\left( {b + c} \right)\left( {b + a} \right)}}{{2b + a + c}} \Leftrightarrow {\left( {a + c} \right)^2} + 2b\left( {c + a} \right) = 2\left( {{b^2} + ab + ac + ab} \right)\\ {a^2} + {c^2} + 2ac + 2bc + 2ba = 2\left( {{b^2} + ab + ac + ab} \right) \Leftrightarrow {a^2} + {c^2} = 2{b^2} \end{array}$