\(ad=bc\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)
\(\Rightarrow\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)
\(\Rightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\rightarrowđpcm\)
ta có : \(ad=bc\Leftrightarrow12ad=12bc\Leftrightarrow6ad+6ad=6bc+6bc\)
\(\Leftrightarrow6ad-6bc=6bc-6ad\Leftrightarrow4ac+6ad-6bc-9bd=4ac+6bc-6ad-9bd\)
\(\Leftrightarrow\left(2a+3b\right)\left(2c-3d\right)=\left(2c+3d\right)\left(2a-3b\right)\Leftrightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\left(đpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Thay vào vế trái ta có:
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
Thay vào vế phải ta có:
\(\dfrac{2c+3d}{2d-3c}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
Từ (1) và (2) \(\Rightarrow\) đpcm