a) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\)
\(\Rightarrow3+\frac{b}{a}=3+\frac{d}{c}\Rightarrow\frac{3a+b}{a}=\frac{3c+d}{c}\)
\(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)
\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{\left(ck\right)^2-\left(dk\right)^2}{c^2-d^2}=k^2\)
và \(\frac{ab}{cd}=\frac{ck.dk}{cd}=k^2\)
Vậy \(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\left(đpcm\right)\)