Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) thì \(a=b.k\) , \(c=d.k\)
Ta tính giá trị của các tỉ số \(\dfrac{a-b}{a};\dfrac{c-d}{c}\) theo \(k\)
\(\dfrac{a-b}{a}=\dfrac{b.k-b}{b.k}=\dfrac{b.\left(k-1\right)}{b.k}=\dfrac{k-1}{k}\left(1\right)\)
\(\dfrac{c-d}{c}=\dfrac{d.k-d}{d.k}=\dfrac{d\left(k-1\right)}{d.k}=\dfrac{k-1}{k}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\) suy ra \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\b=ck\end{matrix}\right.\)
Ta có : \(\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{k}=\dfrac{k-1}{k}\left(1\right)\)
\(\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\dfrac{a-b}{a}=k=\dfrac{c-d}{c}\)
\(\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\left(ĐPCM\right)\)
Vậy \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(1-\dfrac{b}{a}=1-\dfrac{d}{c}\Leftarrow\left\{{}\begin{matrix}\dfrac{c-d}{c}=1-\dfrac{d}{c}\\\dfrac{a-b}{a}=1-\dfrac{b}{a}\\\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\end{matrix}\right.\)
Suy ra \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
Vậy: Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)có thể suy ra\(\dfrac{a-b}{a}=\dfrac{c-d}{c}\).
