Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk; c=dk
\(\dfrac{2a-6b}{2b}=\dfrac{2\cdot bk-6b}{2b}=\dfrac{2b\left(k-3\right)}{2b}=k-3\)
\(\dfrac{2c-6d}{2d}=\dfrac{2\cdot dk-6d}{2d}=\dfrac{2d\left(k-3\right)}{2d}=k-3\)
Do đó: \(\dfrac{2a-6b}{2b}=\dfrac{2c-6d}{2d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Thay \(a=bk;c=dk\) vào\(\dfrac{2a-6b}{2b}=\dfrac{2c-6d}{2d}\) ta được:
\(\dfrac{2a-6b}{2b}=\dfrac{2c-6d}{2d}=\dfrac{2bk-6b}{2b}=\dfrac{2dk-6d}{2d}=k-3\)
Vậy \(\dfrac{2a-6b}{2b}=\dfrac{2c-6d}{2d}\)