Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{{b^2} + {c^2}}}{{2c}} = b\cos a - a\cos b\\
\Leftrightarrow \frac{{{b^2} + {c^2}}}{{2c}} = b.\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} - a.\frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\\
\Leftrightarrow \frac{{{b^2} + {c^2}}}{{2c}} = \frac{{{b^2} + {c^2} - {a^2}}}{{2c}} - \frac{{{a^2} + {c^2} - {b^2}}}{{2c}}\\
\Leftrightarrow {b^2} + {c^2} = {b^2} + {c^2} - {a^2} - {a^2} - {c^2} + {b^2}\\
\Leftrightarrow {b^2} + {c^2} = 2{b^2} - 2{a^2}\\
\Leftrightarrow {c^2} = {b^2} - 2{a^2}
\end{array}\)
