(THẾ LÀ BÉ SAI ĐỀ RỒI NHÁ)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
Ta có: \(a=bk;c=dk\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2.k}{d^2.k}=\dfrac{b^2}{d^2}\\\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\end{matrix}\right.\)
Từ đó ta suy ra được :
\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)