Ta đặt \(\frac{a}{b}=\frac{c}{d}=k\). Ta có \(a=bk\)và \(c=dk\)
Ta có : \(\frac{5a+3b}{5c+3d}=\frac{5bk+3b}{5dk+3d}=\frac{b\left(5k+3\right)}{d\left(5k+3\right)}=\frac{b}{d}\)
\(\frac{5a-3b}{5c-3d}=\frac{5bk-3b}{5dk-3d}=\frac{b\left(5k-3\right)}{d\left(5k-3\right)}=\frac{b}{d}\)
\(\Rightarrow\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\Rightarrowđpcm\).
Cách 2 : Ta có : \(\frac{5a+3b}{5c+3d}=\frac{5bk+3b}{5dk+3d}=\frac{b\left(5k+3\right)}{d\left(5k+3\right)}=\frac{b}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{5a}{5c}=\frac{3b}{3d}\)Áp dụng t/c dãy tỉ số bằng nhau, ta có
\(\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\Rightarrowđpcm\)