Question

Cho \(\frac{a}{b}=\frac{c}{d}\ne0.Cmr:\frac{a-b}{a+b}=\frac{c-d}{c+d}\)

Answer 1

Đặt: \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=b.k\) ; \(c=d.k\)

Ta có:

\(\frac{a-b}{a+b}=\frac{b.k-b}{b.k+b}=\frac{b.\left(k-1\right)}{b.\left(k+1\right)}=\frac{k-1}{k+1}\left(1\right)\)

\(\frac{c-d}{c+d}=\frac{d.k-d}{d.k+d}=\frac{d.\left(k+1\right)}{d.\left(k-1\right)}=\frac{k-1}{k+1}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra: \(\frac{a-b}{a+b}=\frac{c-d}{c+d}\)