ĐẶT \(\frac{a}{b}\)= \(\frac{c}{d}\)là k
suy ra a=kb; c=kd
ta có:\(\frac{2a+13b}{3a-7b}\)= \(\frac{2kb+13b}{3kb-7b}\)= \(\frac{b\left(2k+13\right)}{b\left(3k-7b\right)}\)=\(\frac{2k+13}{3k-7}\) (1)
\(\frac{2c+13d}{3c-7d}\)=\(\frac{2kd+13d}{3kd-7d}\)=\(\frac{d\left(2k+13\right)}{d\left(3k-7\right)}\)=\(\frac{2k+13}{3k-7}\) (2)
từ (1) và (2) suy ra \(\frac{2a+13b}{3a-17b}\)=\(\frac{2c+13d}{3c-7d}\)