a, Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=b.k,c=d.k\)
+) \(\frac{5a+3b}{5a-3b}=\frac{5.b.k+3.b}{5.b.k-3.b}=\frac{b.\left(5k+3\right)}{b.\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(1\right)\)
+) \(\frac{5c+3d}{5c-3d}=\frac{5.d.k+3.d}{5.d.k-3.d}=\frac{d.\left(5k+3\right)}{d.\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(2\right)\)
Từ (1) và(2) => ĐPCM