Đáp án:
\[M = \dfrac{9}{2}\]
Giải thích các bước giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\begin{array}{l}
\dfrac{a}{{2b + c}} = \dfrac{b}{{2c + a}} = \dfrac{c}{{2a + b}} = \dfrac{{a + b + c}}{{\left( {2b + c} \right) + \left( {2c + a} \right) + \left( {2a + b} \right)}} = \dfrac{{a + b + c}}{{3\left( {a + b + c} \right)}} = \dfrac{1}{3}\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{a}{{2b + c}} = \dfrac{1}{3}\\
\dfrac{b}{{2c + a}} = \dfrac{1}{3}\\
\dfrac{c}{{2a + b}} = \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3a = 2b + c\\
3b = 2c + a\\
3c = 2a + b
\end{array} \right.\\
\Rightarrow 6a - 3b = 2.\left( {2b + c} \right) - \left( {2c + a} \right)\\
\Leftrightarrow 6a - 3b = 4b - a\\
\Leftrightarrow 7a = 7b\\
\Leftrightarrow a = b\\
3a = 2b + c\\
\Leftrightarrow 3a = 2a + c\\
\Leftrightarrow a = c\\
\Rightarrow a = b = c\\
M = \dfrac{{2b + c}}{{2a}} + \dfrac{{2c + a}}{{2b}} + \dfrac{{2a + b}}{{2c}} = \dfrac{{2a + a}}{{2a}} + \dfrac{{2a + a}}{{2a}} + \dfrac{{2a + a}}{{2a}} = \dfrac{3}{2} + \dfrac{3}{2} + \dfrac{3}{2} = \dfrac{9}{2}
\end{array}\)
Vậy \(M = \dfrac{9}{2}\)
