C1: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow a=bk;c=dk\)
\(\Rightarrow\dfrac{3a-4b}{b}=\dfrac{bk-4b}{b}=\dfrac{b\left(k-4\right)}{b}=k-4\left(1\right)\)
\(\Rightarrow\dfrac{3c-4d}{d}=\dfrac{dk-4d}{d}=\dfrac{d\left(k-4\right)}{d}=k-4\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{3a-4b}{b}=\dfrac{3c-4d}{d}\)