\(a,\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=K\Rightarrow a=b.k\)

                                  \(\Rightarrow c=d.k\)     

\(-Tacó:\frac{5a+3b}{5a-3b}=\frac{5b.k+3b}{5b.k-3b}=\frac{b.\left(5k+3\right)}{b.\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(1\right)\) 

\(-Tacó:\frac{5c+3d}{5c-3d}=\frac{5d.k+3d}{5d.k-3d}=\frac{d.\left(5k+3\right)}{d.\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(2\right)\)

\(Từ\left(1\right),\left(2\right)\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)