已知a=x/y+z b=y/x+z c=z/x+y 则a/1+a + b/1+b + c/1+c=？

1 . a,b ,c 分别带入，然后化简

由a=x/(y+z)
得a+1=x/(y+z)+(y+z)/(y+z)
=(x+y+z)/(y+z)

1/(a+1)=(y+z)/(x+y+z)
同样的方法
b+1=y/(x+z)+(x+z)/(x+z)
=(x+y+z)/(x+z)
1/(b+1)=(x+z)/(x+y+z)

c+1=z/(x+y)+(x+y)/(x+y)
=(x+y+z)/(x+y)
1/(c+1)=(x+y)/(x+y+z)

这样 a/(1+a) + b/(1+b) + c/(1+c)
=[1-1/(a+1)]+[1-1/(b+1)]+[1-1/(c+1)]
=3-[1/(a+1)+1/(b+1)+1/(c+1)]
=3-[(y+z)/(x+y+z)+(x+z)/(x+y+z)+(x+y)/(x+y+z)]
=3-[(y+z)+(x+z)+(x+y)]/(x+y+z)
=3-2(x+y+z)/(x+y+z)
=3-2
=1