Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\text{Khi đó }\frac{5a+7b}{11a-13b}=\frac{5bk+7b}{11bk-13b}=\frac{b\left(5k+7\right)}{b\left(11k-13\right)}=\frac{5k+7}{11k-13}\left(1\right);\)
\(\frac{5c+7d}{11c-13d}=\frac{5dk+7d}{11dk-13d}=\frac{d\left(5k+7\right)}{d\left(11k-13\right)}=\frac{5k+7}{11k-13}\left(2\right)\)
\(\text{Từ }\left(1\right)\text{và }\left(2\right)\Rightarrow\frac{5a+7b}{11a-13b}=\frac{5c+7d}{11c-13d}\left(\text{ĐPCM}\right)\)