若a+b+c=0,且abc≠0,求a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)的值

a{1/b+1/c}+b{1/c+1/a}+c{1/a+1/b}
=a/b+a/c+b/c+b/a+c/a+c/b
=[a(b+c)+b(a+c)+c(a+b)]/abc
=(2ab+2ac+2bc)/abc.①
∵a+b+c=0;
∴(a+b+c)(a+b+c)=a*a+b*b+c*c+2ab+2ac+2bc=0;
∴2ab+2ac+2bc=0
∴①=0/abc
=0
2.(X+1/Y)(Z+1/X)(Y+1/Z)=XYZ+1/(XYZ)+X+Y+Z+1/X+1/Y+1/Z=XYZ+1/(XYZ)+4+1+7/3=4*1*7/3
得到：
XYZ+1/(XYZ)=2,解得XYZ=1