Question

求助数学的：已知a^2+b^2+c^2=1,a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=-3,求a+b+c

顺便讲下方法！谢谢

Answer 1

a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=a/b+a/c+b/a+b/c+c/a+c/b
=(a+b)/c+(b+c)/a+(c+a)/b
=c/c+(a+b)/c+a/a+(b+c)/a+b/b+(c+a)/b-3
=(a+b+c)(1/a+1/b+1/c)-3
则(a+b+c)(1/a+1/b+1/c)=0
a+b+c=0 或 ab+bc+ca=0
当ab+bc+ca=0
则(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=1
则a+b+c=±1
故a+b+c=0,或a+b+c=±1