Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{b}=\frac{2c}{d}\)
Đặt:\(\frac{2a}{b}=\frac{2c}{d}=k\left(k\ne0\right)\)
=> 2a=bk; 2c=dk
Ta có:\(\frac{2a+3b}{2a-3b}=\frac{bk+3b}{bk-3b}=\frac{b\left(k+3\right)}{b\left(k-3\right)}=\frac{k+3}{k-3}\left(1\right)\)
\(\frac{2c+3d}{2c-3d}=\frac{dk+3d}{dk-3d}=\frac{d\left(k+3\right)}{d\left(k-3\right)}=\frac{k+3}{k-3}\left(2\right)\)
Từ \(\left(1\right)và\left(2\right)\Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
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