Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
\(\Rightarrow\frac{3a-4c}{3b-4d}=\frac{3bk-4dk}{3b-4d}=\frac{k.\left(3b-4d\right)}{3b-4d}=k\)(1)
\(\frac{5a-6c}{5b-6d}=\frac{5bk-6dk}{5b-6d}=\frac{k.\left(5b-6d\right)}{5b-6d}=k\)(2)
Từ (1) và (2)
\(\Rightarrow\frac{3a-4c}{3b-4d}=\frac{5a-6c}{5b-6d}\)
đpcm
Cách 2:
ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{3a}{3b}=\frac{5a}{5b}=\frac{4c}{4d}=\frac{6c}{6d}\) (*)
\(\Rightarrow\frac{3a}{3b}=\frac{4c}{4d}=\frac{3a-4c}{3b-4d}\)(**)
\(\frac{5a}{5b}=\frac{6c}{6d}=\frac{5a-6c}{5b-6d}\)(***)
Từ (*) ;(**);(***) \(\Rightarrow\frac{3a-4c}{3b-4d}=\frac{5a-6c}{5b-6d}\left(đpcm\right)\)
