Lời giải:
Đặt `a/b=c/d=k-> a=bk, c=dk`
`a)`
`(a-b)/(2a)=(bk-b)/(2bk)=(b(k-1))/(2bk)=(k-1)/(2k)`
`(c-d)/(2c)=(dk-d)/(2dk)=(d(k-1))/(2dk)=(k-1)/(2k)`
`-> (a-b)/(2a)=(c-d)/(2c)`
` ----`
`(a+b)/b=(bk+b)/b=(b(k+1))/b=k+1`
` (c+d)/d=(dk+b)/d=(d(k+1))/d=k+1`
`-> (a+b)/b=(c+d)/d`
`b)`
`(5a-3b)/(3a+2b)=(5bk-3b)/(3bk+2b)=(5k-3)/(3k+2)`
`(5c-3d)/(3c+2d)=(5dk-3d)/(3dk+2d)=(5k-3)/(3k+2)`
`-> (5a-3b)/(3a+2b)=(5c-3d)/(3c+2d)`
