Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\left(1\right)\)
\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
từ \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=>\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=k=>a=ck và b=dk
tacó\(\dfrac{a-b}{a+b}=\dfrac{ck-dk}{ck+dk}=\dfrac{k.\left(c-d\right)}{k.\left(c+d\right)}=\dfrac{c-d}{c+d}\)
=>\(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\) (đpcm)
