已知x/y+z=a,y/x+z=b.z/x+y=c且abc不等于0求a+1分之a+b+1分之b+c+1分之c 等于多少
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![已知x/y+z=a,y/x+z=b.z/x+y=c且abc不等于0求a+1分之a+b+1分之b+c+1分之c 等于多少](/uploads/image/z/5202074-2-4.jpg?t=%E5%B7%B2%E7%9F%A5x%2Fy%2Bz%3Da%2Cy%2Fx%2Bz%3Db.z%2Fx%2By%3Dc%E4%B8%94abc%E4%B8%8D%E7%AD%89%E4%BA%8E0%E6%B1%82a%2B1%E5%88%86%E4%B9%8Ba%2Bb%2B1%E5%88%86%E4%B9%8Bb%2Bc%2B1%E5%88%86%E4%B9%8Bc+%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91)
已知x/y+z=a,y/x+z=b.z/x+y=c且abc不等于0求a+1分之a+b+1分之b+c+1分之c 等于多少
已知x/y+z=a,y/x+z=b.z/x+y=c且abc不等于0求a+1分之a+b+1分之b+c+1分之c 等于多少
已知x/y+z=a,y/x+z=b.z/x+y=c且abc不等于0求a+1分之a+b+1分之b+c+1分之c 等于多少
这种东西叫轮换式.它的意思是x换成y,y换成z,z换成x时,值不变.
求a/(a+1)+a/(b+1)+c/(c+1)
怎么求它呢?
a/a+1=1- 1/(a+1)
b/b+1=1-1/(b+1)
c/c+1=1-1/(c+1)
a/(a+1)+a/(b+1)+c/(c+1)=3-1/(1+a)-1/(1+b)-1/(1+c)①
1+a=1+x/y+z=(x+y+z)/(y+z)
1/(1+a)=(y+z)/(x+y+z)
①变为3-(y+z)/(x+y+z)-(y+x)/(x+y+z)-(x+z)/(x+y+z)
=3-2(x+y+z)/(x+y+z)
=1
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