Ta có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\Leftrightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\Leftrightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)( đpcm )
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\frac{a+b}{c+d}=\frac{bk+b}{dk+d}=\frac{b.\left(k+1\right)}{d.\left(k+1\right)}=\frac{b}{d}\left(1\right)\)
\(\frac{a-b}{c-d}=\frac{bk-b}{dk-d}=\frac{b.\left(k-1\right)}{d.\left(k+1\right)}=\frac{b}{d}\left(2\right)\)
Từ ( 1 ) và ( 2 ) => \(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)hay \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)( đpcm )