Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => a = bk ; c = dk
\(\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bk.dk}{\left(dk\right)^2-bk.dk}=\frac{b^2.k^2+k^2bd}{d^2k^2-k^2bd}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}\) (đpcm)
Vậy \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)