Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) (a ≠ c; b ≠ d)
CMR: \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\)
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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{a+c}{a-c}=\dfrac{bk+dk}{bk=dk}=\dfrac{\left(b+d\right)k}{\left(b-d\right)k}=\dfrac{b+d}{b-d}\)
Mà \(\dfrac{b+d}{b-d}=\dfrac{b+d}{b-d}\)
=> \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\) (đpcm)