Nếu điều cần chứng minh đúng thì:
\(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)
Khi đó:
\(\dfrac{a+b}{a-b}=\dfrac{bt+b}{bt-b}=\dfrac{b\left(t+1\right)}{b\left(t-1\right)}=\dfrac{t+1}{t-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dt+d}{dt-d}=\dfrac{d\left(t+1\right)}{d\left(t-1\right)}=\dfrac{t+1}{t-1}\)
Nên \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)(đúng với điều đề bài cho) nên \(\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Leftrightarrow\left(a+b\right)\left(c-d\right)=\left(a-b\right)\left(c+d\right)\)
\(\Leftrightarrow a\left(c-d\right)+b\left(c-d\right)=a\left(c+d\right)+b\left(c+d\right)\)
\(\Leftrightarrow ac-ad+bc-bd=ac+ad-bc+bd\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\dfrac{a}{d}=\dfrac{b}{c}\rightarrowđpcm\)
Chúc bạn học tốt!
