Đăt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) →\(\left\{{}\begin{matrix}a=bk\\c=dk\\\end{matrix}\right.\)
→\(\dfrac{a+c}{b+d }=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\)
→ \(\dfrac{2a+3c}{2b+3d}=\dfrac{2.bk+3.dk}{2b+3d}=k\)
Vậy \(\dfrac{a+c}{b+d}=\dfrac{2a+3.dk}{2b+3d}\left(=k\right)\).