Giải:
a=b+c; $c=\frac{bd}{b-d}$
$c=\frac{bd}{b-d}$\Leftrightarrow$\frac{1}{c}=\frac{b-d}{bd}=\frac{1}{d}-\frac{1}{b}$
Nhân cả hai vế với c, ta được:
$\frac{c}{c}=\frac{c}{d}-\frac{c}{b}$
\Leftrightarrow$1+\frac{c}{b}=\frac{c}{d}$
\Leftrightarrow$\frac{b+c}{b}=\frac{c}{d}$
mà a=b+c\Rightarrow[TEX]\frac{a}{b}=\frac{c}{d}[/TEX]
Ta có c=$\dfrac{bd}{b-d}$\Rightarrow$\dfrac{1}{c}=\dfrac{b-d}{bd}=\dfrac{b}{bd}-\dfrac{d}{bd}=\dfrac{1}{d}-\dfrac{1}{b}$
\Rightarrow$\dfrac{c}{c}=\dfrac{c}{d}-\dfrac{c}{b}$
\Rightarrow1+$\dfrac{c}{b}=\dfrac{c}{b}$
=$\dfrac{b+c}{b}=\dfrac{c}{b}$ hay$\dfrac{a}{b}=\dfrac{c}{d}$