已知abc不等于0,a+b+c=1,a^2+b^2+c^2=1,求(2b+2c/a)+(2a+2b/b)+(2a+2b/

题目应该是"(2b+2c)/a+(2a+2c)/b+(2a+2b)/c"吧!
(2b+2c)/a+(2a+2c)/b+(2a+2b)/c
=2(b+c)/a+2(a+c)/b+2(a+b)/c
=2(1-a)/a+2(1-b)/b+2(1-c)/c
=2(1/a+1/b+1/c)-6
=2[(bc+ac+ab)/(abc)]-6,
因为(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc,
所以1=1+2(ab+ac+bc),
所以ab+bc+ac=0,
所以原式=2[(bc+ac+ab)/(abc)]-6
=2*0-6
=-6.