Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\widehat A = 60^\circ \\
{m_b} = 8\\
b + c = a\sqrt 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\
{m_b}^2 = \frac{{{a^2} + {c^2}}}{2} - \frac{{{b^2}}}{4}\\
b + c = a\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{b^2} + {c^2} - {a^2} = bc\\
\frac{{{a^2} + {c^2}}}{2} - \frac{{{b^2}}}{4} = 64\\
b + c = a\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{b^2} + {c^2} - {\left( {\frac{{b + c}}{{\sqrt 3 }}} \right)^2} = bc\\
2\left[ {{{\left( {\frac{{b + c}}{{\sqrt 3 }}} \right)}^2} + {c^2}} \right] - {b^2} = 64.4\\
b + c = a\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{b^2} + {c^2} - \frac{{{b^2} + 2bc + {c^2}}}{3} = bc\\
2.\left[ {\frac{{{b^2} + {c^2} + 2bc}}{3} + {c^2}} \right] - {b^2} = 256\\
b + c = a\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2{b^2} - 5bc + 2{c^2} = 0\\
- \frac{1}{3}{b^2} + \frac{4}{3}bc + \frac{5}{3}{c^2} = 256\\
b + c = a\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
b = 2c\\
b = \frac{1}{2}c
\end{array} \right.\\
- \frac{1}{3}{b^2} + \frac{4}{3}bc + \frac{5}{3}{c^2} = 256\\
b + c = a\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
b = 2c\\
3{c^2} = 256\\
b + c = a\sqrt 3
\end{array} \right.\\
\left\{ \begin{array}{l}
c = 2b\\
9{b^2} = 256\\
b + c = a\sqrt 3
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
c = \frac{{16\sqrt 3 }}{3}\\
b = \frac{{32\sqrt 3 }}{3}\\
a = 16
\end{array} \right.\\
\left\{ \begin{array}{l}
b = \frac{{16}}{3}\\
c = \frac{{32}}{3}\\
a = \frac{{16\sqrt 3 }}{3}
\end{array} \right.
\end{array} \right.
\end{array}\)
