1,xy/x+y=a,xz/x+z=b,yz/y+z=c
分式倒过来,
(x+y)/(xy)=1/a,(x+z)/(xz)=1/b,(y+z)/(yz)=1/c
1/x+1/y=1/a,1/x+1/z=1/b,1/y+1/z=1/c
3式相加得:
1/x+1/y+1/z=(1/a+1/b+1/c)/2
1/x=(1/a+1/b+1/c)/2-1/c=1/2a+1/2b-1/2c
x=1/(1/2a+1/2b-1/2c)=2abc/(bc+ac-ab)
2）若a+b+c=0,abc不=0,求(a^2+b^2+c^2/a^3+b^3+c^3)+2/3(1/a+1/b+1/c)的值
a+b+c=0.
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0
a^3+b^3+c^3=3abc
(a^2+b^2+c^2/a^3+b^3+c^3)+2/3(1/a+1/b+1/c)
=(a^2+b^2+c^2)/(a^3+b^3+c^3)+2(ab+bc+ca)/(3abc)
=(a^2+b^2+c^2+2ab+2bc+2ca)/(3abc)
=(a+b+c)^2/(3abc)
=0