Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
\(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b\left(2k+13\right)}{b\left(3k-7\right)}=\frac{2k+13}{3k-7}\left(1\right)\)
\(\frac{2c+13d}{3c-7d}=\frac{2dk+13d}{3dk-7d}=\frac{d\left(2k+13\right)}{d\left(3k-7\right)}=\frac{2k+13}{3k-7}\left(2\right)\)
Từ \(\left(1\right)\) và (2) \(\Rightarrow\frac{a}{b}=\frac{c}{d}\)( đpcm )
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Từ \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)\(\Rightarrow\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}=\frac{2a}{2c}=\frac{13b}{13d}=\frac{3a}{3c}=\frac{7b}{7d}=\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)