一道分式题若x/(y+z+t)=
假设x/(y+z+t)=y/(x+z+t)=z/(x+y+z)=t/(x+y+z)=1/k
--->y+z+t=kx; x+z+t=ky; x+y+t=kz; x+y+z=kt
把此4等式的两边分别相加得到
3(x+y+z+t)=k(x+y+z+t)--->1,x+y+z+t=0 or 2,k=3
1,:x+y=-(z+t)--->(x+y)/(z+t)=-1,
also (y+z)/(x+t)=-1,(z+t)/(x+y)=-1,(t+x)/(z+y)=-1。
所以,f=-1-1-1-1=-4。
2,k=3--->x+y+z=3t & x+z+t=3x
--->x-y=3(y-x...全部
假设x/(y+z+t)=y/(x+z+t)=z/(x+y+z)=t/(x+y+z)=1/k
--->y+z+t=kx; x+z+t=ky; x+y+t=kz; x+y+z=kt
把此4等式的两边分别相加得到
3(x+y+z+t)=k(x+y+z+t)--->1,x+y+z+t=0 or 2,k=3
1,:x+y=-(z+t)--->(x+y)/(z+t)=-1,
also (y+z)/(x+t)=-1,(z+t)/(x+y)=-1,(t+x)/(z+y)=-1。
所以,f=-1-1-1-1=-4。
2,k=3--->x+y+z=3t & x+z+t=3x
--->x-y=3(y-x)
--->x-y=0
--->x=y also y=z z=t
--->x=y=z=t=m<>0
--->x+y=2m y+z=2m z+t=2m t+x=2m
所以 f=1+1+1+1=4
所以,f的值是+'-4,是整数。收起
