đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k => a=bk; b=dk
\(\dfrac{a}{3b}\)=\(\dfrac{bk}{3b}=\dfrac{k}{3}\)
\(\dfrac{c}{3d}=\)\(\dfrac{dk}{3d}=\)\(\dfrac{k}{3}\)
=>\(\dfrac{a}{3b}=\)\(\dfrac{c}{3d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
ta có : \(\dfrac{a}{3b}=\dfrac{bk}{3b}=\dfrac{k}{3}\left(1\right)\)
lại có \(\dfrac{c}{3d}=\dfrac{dk}{3d}=\dfrac{k}{3}\left(2\right)\)
từ (1) và (2) => \(\dfrac{a}{3b}=\dfrac{c}{3d}\)