Đặt $\dfrac ab =\dfrac cd = k$
$\to \begin{cases}a = kb\\c = kd\end{cases}$
Ta có:
$+)\quad \dfrac{ab}{cd}=\dfrac{kb^2}{kd^2}=\dfrac{b^2}{d^2}$
$+)\quad\dfrac{a^2 - b^2}{c^2 - d^2}=\dfrac{k^2b^2 - b^2}{k^2d^2 - d^2}$
$=\dfrac{(k^2 -1)b^2}{(k^2-1)d^2}$
$=\dfrac{b^2}{d^2}$
Vậy $\dfrac{ab}{cd}=\dfrac{a^2 - b^2}{c^2 - d^2}$