1、已知abc≠0,且a+b+c=0,则代数式a^2/bc+b^2/ac+c^2/ab的值为_____

1.
a^2/bc+b^2/ac+c^2/ab=(a^3+b^3+c^3)/abc
=[(a+b)^3-3ab^2-3ba^2+c^3]/abc
=[(-c)^3-3ab^2-3ba^2+c^3]/abc
=-3ab(a+b)/abc
=-3(-c)/c
=3
2
x+2/(x-1)=a+2/(a-1)等价于(x-1)++2/(x-1)=(a-1)+2/(a-1)
设x-1=x`,a-1=c`,x=x`+1
则原式等价于x`+2/x`=c`+2/c`,它的两个解是x1`=c`,x2`=2/c`
所以原式解为x1=c`+1=a-1+1=a,x2=2/c`+1=1+2/(a-1)=(a+1)/(a-1)

1.检举a+b+c=0,a+b=-c
a^2/bc+b^2/ac+c^2/ab
通分
=(a^3+b^3+c^3)/abc
=[(a+b)(a^2-ab+b^2)+c^3]/abc
=[-c(a^2-ab+b^2)+c^3]/abc
={-c[(a+b)^2-3ab]+c^3}/abc
=[-c(c^2-3ab)+c^3]/abc
=(-c^2+3abc+c^3)/abc
=3abc/abc
=3

1.因为a+b+c=0，则有a=-(b+c),则有a^2=[-(b+c)]^2=b^2+c^2+2bc
则a^2/bc=c/b+b/c+2 同理可得原式等于
c/b+b/c+2 + c/a+a/c+2 + b/a+a/b+2 =
(b+c)/a + (a+c)/b +(a+b)/c +6
因为a+b...