Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{a-b}{b}=\dfrac{bk-b}{b}=\dfrac{b\left(k-1\right)}{b}=k-1\left(1\right)\)
\(VP=\dfrac{c-d}{d}=\dfrac{dk-d}{d}=\dfrac{d\left(k-1\right)}{d}=k-1\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\\ \Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\\ \Leftrightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\left(đpcm\right)\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
\(\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\left(đpcm\right)\)
Chúc bạn học tốt!