英语数学（分式）(1111:26:51)已知非零实数abc满足a^2+b^2+c^2=1,且a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值.

Question

英语 数学（分式） (11 11:26:51)
已知非零实数abc满足a^2+b^2+c^2=1,且a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值.

Answer 1

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
a(1/b+1/c)+1+b(1/c+1/a)+1+c(1/a+1/b)+1=-3+3
a(1/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=0
(a+b+c)*(1/a+1/b+1/c)=0
a+b+c=0
或1/a+1/b+1/c=0
(bc+ac+ab)/(abc)=0
ab+ac+bc=0
a^2+b^2+c^2=1
a^2+b^2+c^2+2ab+2ac+2bc=1+0
(a+b+c)^2=1
a+b+c=1或-1
综上所述a+b+c=0或1或-1