Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=> a = b.k ; c= d.k
\(\frac{2a+3b}{2a-3b}=\frac{2.\left(b.k\right)+3.b}{2.\left(b.k\right)-3b}=\frac{2b.k+3b}{2b.k-3b}=\frac{2b.\left(k+1,5\right)}{2b.\left(k-1,5\right)}=\frac{k+1,5}{k-1,5}\left(1\right)\)
\(\frac{2c+3d}{2c-3d}=\frac{2.\left(d.k\right)+3d}{2.\left(d.k\right)-3d}=\frac{2d.k+3d}{2d.k-3d}=\frac{2d.\left(k+1,5\right)}{2d.\left(k-1,5\right)}=\frac{k+1,5}{k-1,5}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) => đpcm
a/c=c/d suy ra 2a/2c=3b/3d = (2a+3b)/(2c+3d)=(2a-3d)/(2c-3d)
suy ra (2a+3b)/(2a-3b)=(2c+3d)/(2c-3d)
vì a=c và b=d nên bài này dễ mà