Bài này có 4 cách nhé mik làm luôn cả 4 cách
Cách 1: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\)
Xét tích (a-b).c=ac-bc=ac-ad=a.(c-d)
Vay (a-b).c =a.(c-d)
=>\(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
Cách 2:
Ta đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=kb;c=kd\)
Thế thì \(\dfrac{a-b}{a}=\dfrac{kb-b}{kb}=\dfrac{b.\left(k-1\right)}{kb}=\dfrac{k-1}{k}\) (1)
\(\dfrac{c-d}{c}=\dfrac{kd-d}{kd}=\dfrac{d.\left(k-1\right)}{kd}=\dfrac{k-1}{k}\) (2)
Từ (1),(2) => \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
Cách 3:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)
Vay \(\dfrac{a-b}{c-d}=\dfrac{a}{c}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
Cách 4:
Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)nen \(\dfrac{b}{a}=\dfrac{d}{c}\)
Ta có: \(\dfrac{a-b}{a}=\dfrac{a}{a}-\dfrac{b}{a}=1-\dfrac{b}{a}=1-\dfrac{d}{c}=\dfrac{c-d}{c}\)
Vậy \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
mik thiếu đề bài nhé
CMR: \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
Vậy nhé
