Nếu \(\frac{1}{{b + c}};\,...

D

- Hướng dẫn giải

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

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