八年级下分式化简题。

显示全部楼层 · 2013-8-31 14:27:26

a+b+c=0所以a+b=-ca+c=-bb+c=-a所以2[a*(1/b+1/c)+b*(1/c+1/a)+c*(1/a+1/b)]=2[a/b+a/c+b/c+b/a+c/a+c/b]=2[(b+c)/a+(a+c)/b+(a+b)/c]=2[(-a)/a+(-b)/b+(-c)/c]=2（-1-1-1）=-6...

千问 · 2013-8-31 14:27:26

c(2/a+2/b)+b(2/a+2/c)+a(2/b+2/c)=2c/a+2c/b+2b/a+2b/c+2a/b+2a/c=2[(c+b)/a+(c+a)/b+(a+b)/c]=2[-a/a-b/b-c/c]=-6...