Đáp án: $\dfrac{13}{3}$
Giải thích các bước giải:
Trường hợp $1: a+b+c=0$
$\to 2a+b=a+(a+b)=a-c$
$2b+c=b+(b+c)=b-a$
$2c+a=c+(c+a)=c-b$
Mà $\dfrac{2a+b}{c}=\dfrac{2b+c}{a}=\dfrac{2c+a}{b}$
$\to \dfrac{a-c}{c}=\dfrac{b-a}{a}=\dfrac{c-b}{b}$
$\to \dfrac{a}{c}-1=\dfrac{b}{a}-1=\dfrac{c}{b}-1$
$\to \dfrac{a}{c}=\dfrac{b}{a}=\dfrac{c}{b}$
$\to (\dfrac{a}{c})^3=(\dfrac{b}{a})^3=(\dfrac{c}{b})^3=\dfrac{a}{c}.\dfrac{b}{a}.\dfrac{c}{b}=1$
$\to \dfrac{a}{c}=\dfrac{b}{a}=\dfrac{c}{b}=1$
$\to a=b=c$
Lại có $a+b+c=0\to a =b=c=0$ loại
$\to a+b+c\ne 0$
Ta có:
$\dfrac{2a+b}{c}=\dfrac{2b+c}{a}=\dfrac{2c+a}{b}=\dfrac{2a+b+2b+c+2c+a}{c+a+b}=\dfrac{3(a+b+c)}{a+b+c}=3$
$\to\begin{cases} 2a+b=3c\\ 2b+c=3a\\ 2c+a=3b\end{cases}$
$\to \dfrac{2a+b}{c}+\dfrac{a}{2b+c}+\dfrac{3b}{2c+a}=\dfrac{3c}{c}+\dfrac{a}{3a}+\dfrac{3b}{3b}=\dfrac{13}{3}$
