Đặt \(\frac{a}{2}=\frac{c}{4}=\frac{e}{5}\) = k => a = 2k; c = 4k ; e = 5k
\(\frac{b}{3}=\frac{d}{5}=\frac{g}{6}\)= h => b = 3h; d = 5h; g = 6h
Khi đó: \(\frac{a}{b}+\frac{c}{d}+\frac{e}{g}=\frac{2k}{3h}+\frac{4k}{5h}+\frac{5k}{6h}=\left(\frac{2}{3}+\frac{4}{5}+\frac{5}{6}\right).\frac{k}{h}=2\frac{3}{10}\)
=> \(\frac{23}{10}.\frac{k}{h}=\frac{23}{10}\)=> \(\frac{k}{h}=1\)=> k = h
Vậy \(\frac{a}{b}=\frac{2k}{3h}=\frac{2}{3};\frac{c}{d}=\frac{4}{5};\frac{e}{g}=\frac{5}{6}\)