Ta có:
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow ad=bc\)
\(\Rightarrow ac-ad=ac-bc\)
\(\Rightarrow a\left(c-d\right)=c\left(a-b\right)\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
Vậy \(\frac{a}{a-b}=\frac{c}{c-d}\)
Ta có :
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow ad=bc\)
\(\Rightarrow ac-ad=ac-bc\)
\(\Rightarrow a\left(c-d\right)=c\left(a-b\right)\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
\(KL:\frac{a}{a-b}=\frac{c}{c-d}\)
\(\text{Ta có : }\frac{a}{b}=\frac{c}{d}\left(a,b,c\ne0;a\ne b\ne c\ne d\right)\)
\(\Rightarrow ad=cb\left(\text{tính chất tỉ lệ thức}\right)\)
\(\Rightarrow ac-ad=ac-cb\left(\text{tính chất của đẳng thức}\right)\)
\(\Rightarrow a\left(c-d\right)=c\left(a-b\right)\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\left(đpcm\right)\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k,k\inℕ^∗\)
\(\Rightarrow a=bk,c=dk\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
\(\Leftrightarrow\frac{a}{bk-b}=\frac{c}{dk-d}\)
\(\Leftrightarrow\frac{bk}{b\left(k-1\right)}=\frac{dk}{d\left(k-1\right)}\)
\(\Leftrightarrow\frac{k}{k-1}=\frac{k}{k-1}\)(luôn đúng)
vậy.......