đặt : \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=b.k;c=d.k\)
Ta có :\(\frac{2a-3b}{2a+3b}=\frac{2.b.k-3b}{2.b.k+3b}=\frac{b\left(2k-3\right)}{b\left(2k+3\right)}=\frac{2k-3}{2k+3}\)
\(=\) \(\frac{d\left(2k-3\right)}{d\left(2k+3\right)}=\frac{2d.k-3d}{2.d.k+3d}=\frac{2c-3d}{2c+3d}\left(c=d.k\right)\)
Đặt a / b = c/d = K . Suy ra :
a = b.K ; c = d.K
Ta có: 2a-3b/2a+3b = 2.b.K-3b/2.b.K+3b =b(2.k-3)/ b(2.k+3) = 2k-3/2k+3 (1)
2c-3d/2c+3b = 2.d.k-3d/2.d.k+3d = d(2.k-3)/d(2.k+3) = 2k-3/2k+3 (2)
Từ (1) và (2) . Suy ra : 2a-3b/2a+3b = 2c-3d/2c+3d