Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Suy ra: \(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b.\left(2k+13\right)}{b.\left(3k-7\right)}=\frac{2k+13}{3k-7}\)
\(\frac{2c+13d}{3c-7d}=\frac{2dk+13d}{3dk-7d}=\frac{d.\left(2k+13\right)}{d.\left(3k-7\right)}=\frac{2k+13}{3k-7}\)
Vậy \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\) khi: \(\frac{a}{b}=\frac{c}{d}\)