若x/(y+z+t)=y/(z+t+x)=z/(t+x+y)=t/(x+y+z)即f=(x+y)/(z+t)+(y+z)/(x+t)+(z+t)/(x+y)=(t+x)/(y+z)证明：f是一个整数f=(x+y)/(z+t)+(y+z)/(x+t)+(z+t)/(x+y)+(t+x)/(y+z)后面那个是加号，不好意思，打错了

Question

若 x/(y+z+t)=y/(z+t+x)=z/(t+x+y)=t/(x+y+z)
即 f=(x+y)/(z+t)+(y+z)/(x+t)+(z+t)/(x+y)=(t+x)/(y+z)
证明：f是一个整数
f=(x+y)/(z+t)+(y+z)/(x+t)+(z+t)/(x+y)+(t+x)/(y+z)
后面那个是加号，不好意思，打错了

Answer 1

题目中f的表达式中最后那个等号应该是+号吧
f=(x+y)/(z+t)+(y+z)/(x+t)+(z+t)/(x+y)+(t+x)/(y+z)
设：a = x + y + z + t
并设：
k = x/(y+z+t) = y/(z+t+x) = z/(t+x+y) = t/(x+y+z)
由k = x/(y+z+t),得：
k = x/(a-x)
即：
(k+1)x = ak ………………①
同理可得：
(k+1)y = ak ………………②
(k+1)z = ak ………………③
(k+1)t = ak ………………④
（1）若k = -1,则由①得：
0*x = -a
得：a = 0,即x + y + z + t = 0
即：x+y = -(z+t)；y+z = -(x+t)；z+t = -(x+y)；t+x = -(y+z)
因此:
f = (x+y)/(z+t) + (y+z)/(x+t) + (z+t)/(x+y) + (t+x)/(y+z)
= -1*4 = -4
（2）若k ≠ -1,则由①~④得：
x = y = z = t = ak/(k+1)
因此：
f = (x+y)/(z+t) + (y+z)/(x+t) + (z+t)/(x+y) + (t+x)/(y+z)
= 1*4 = 4
由（1）（2）知,f为整数