[anongc]

$\frac{a}{b+c}=\frac{2}{9}$
$\frac{b}{c+a}=\frac{3}{8}$
$\frac{c}{a+b}=\frac{6}{5}$
$จงหา a+b+c ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
$a+b+c ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ )=1+\frac{b+c}{a}+1+\frac{a+c}{b}+1+\frac{a+b}{c}$
$=3+\frac{9}{2}+\frac{8}{3}+\frac{5}{6}$
$=11$