Lời giải:
Áp dụng TCDTSBN:
$a+b+c=\frac{a+2b-c}{c}=\frac{b+2c-a}{a}=\frac{c+2a-b}{b}=\frac{a+2b-c+b+2c-a+c+2a-b}{c+a+b}=\frac{2(a+b+c)}{a+b+c}=2$
\(\Rightarrow \left\{\begin{matrix} a+b+c=2\\ a+2b-c=2c\\ b+2c-a=2a\\ c+2a-b=2b\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a+b+c=2\\ a+2b=3c\\ b+2c=3a\\ c+2a=3b\\ \end{matrix}\right.\)
Có:
$a+2b=3c=3(2-a-b)=6-3a-3b$
$\Rightarrow 4a+5b=6(1)$
$b+2c=3a$
$\Rightarrow b+2(2-a-b)=3a$
$\Rightarrow 4-2a-b=3a$
$\Rightarrow 5a+b=4(2)$
Từ $(1); (2)\Rightarrow 5(5a+b)-(4a+5b)=14$
$\Rightarrow 21b=14\Rightarrow b=\frac{2}{3}$
$5a=4-b=4-\frac{2}{3}=\frac{10}{3}\Rightarrow a=\frac{2}{3}$
$c=2-a-b=2-\frac{2}{3}-\frac{2}{3}=\frac{2}{3}$
