Question

cho : \(\frac{a}{b}=\frac{c}{d}\) \(\left(a;b;c\ne0;a\ne b,b\ne c;c\ne d\right)\) 

c\m : \(\frac{a}{a-b}=\frac{c}{c-d}\)

Answer 1

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 \(\frac{a}{a-b}=\frac{c}{c-d}\)

Answer 2

Ta có :

\(\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow ad=bc\)

\(\Rightarrow ac-ad=ac-bc\)

\(\Rightarrow a\left(c-d\right)=c\left(a-b\right)\)

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

\(KL:\frac{a}{a-b}=\frac{c}{c-d}\)

Answer 3

\(\text{Ta có : }\frac{a}{b}=\frac{c}{d}\left(a,b,c\ne0;a\ne b\ne c\ne d\right)\)

\(\Rightarrow ad=cb\left(\text{tính chất tỉ lệ thức}\right)\)

\(\Rightarrow ac-ad=ac-cb\left(\text{tính chất của đẳng thức}\right)\)

\(\Rightarrow a\left(c-d\right)=c\left(a-b\right)\)

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

Answer 4

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.......