Question

Từ đẳng thức \(\frac{a}{b}=\frac{c}{d}\) hãy suy ra tỉ lệ \(\frac{a}{a-b}=\frac{c}{c-d}\)

Answer 1

Đặt :

\(\frac{a}{b}=\frac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có :

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

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

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\frac{a}{a-b}=\frac{c}{c-d}\left(đpcm\right)\)

Answer 2

Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

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

\(\Rightarrow\frac{b}{a}-1=\frac{d}{c}-1.\)

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

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

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

Chúc bạn học tốt!

Answer 3

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) thì a=bk,c=dk.

Ta có:

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

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

Từ (1) và (2) ⇒ \(\frac{a}{a-b}=\frac{c}{c-d}\left(\text{đ}pcm\right)\)

Vậy \(\frac{a}{a-b}=\frac{c}{c-d}\)

Answer 4

Mình ko chép đâu đấy