Question

Chứng minh rằng: nếu \(\frac{a}{b}=\frac{c}{d}\ne1\) thì \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\) với \(a,b,c,d\ne0\)

Answer 1

Với \(a,b,c,d\ne0\) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\)

\(\Rightarrow\frac{a+b}{c+d}=\frac{b}{d}\left(1\right)\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\Rightarrow\frac{a-b}{c-d}=\frac{b}{d}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)