$\\$
Đặt `a/b=c/d=k(k\ne 0)`
`=>a=bk,c=dk`
`(2015a+2016b)/(2015a - 2016b)`
`= (2015 bk + 2016b)/(2015bk - 2016b)`
`= (b(2015k + 2016))/(b (2015k - 2016))`
`= (2015k+2016)/(2015k - 2016) (1)`
`(2015c + 2016d)/(2015c-2016d)`
`= (2015dk + 2016d)/(2015dk - 2016d)`
`= (d (2015k + 2016))/(d(2015k - 2016))`
`= (2015k + 2016)/(2015k - 2016)(2)`
`(1)(2) => (2015a+2016b)/(2015a - 2016b)=(2015c + 2016d)/(2015c-2016d)`