(+) \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)

(+) \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (*)

\(\Leftrightarrow4ac+6bc-6ad-9bd=4ac-6bc+6ad-9bd\)

\(\Leftrightarrow12bc=12ad\Leftrightarrow bc=ad\) (đúng)

Vậy (*) đúng (đpcm)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\)(đpcm)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\left(1\right)\)

\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3b}\left(=\dfrac{2k+3}{2k-3}\right)\)

Áp dụng tính chất dãy tỉ số băng nhau,ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{2a}{2c}=\dfrac{3b}{3d}=>\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3d}{2c-3d}=>\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\left(đpcm\right)\)

Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a}{c}=\dfrac{b}{d}\)

=> \(\dfrac{2a}{2c}=\dfrac{3b}{3d}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{2a}{2c}=\dfrac{3b}{3d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{3c-3d}\)

Vậy \(\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\) (ĐPCM)