Đặt `a/b = c/d = k`

`=> {(a=bk),(c=dk):}`

Ta có :

`( 2a + 5b)/(3a-4b) = (2bk + 5b)/(3bk-4b)=[b(2k+5)]/[b(3k-4)] = (2k+5)/(3k-4)`

`(2c+5d)/(3c-4d) = (2dk+5d)/(3dk-4d) = [d(2k+5)]/[d(3k-4)] = (2k+5)/(3k-4`

`=> (2a+5b)/(3a-4b) = (2c+5d)/(3c-4d)`