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

题目

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
1.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
2.已知x、y、z满足x+1/y=4,y+1/z=1,z+1/x=7/3,求xyz的值.

答案

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 =02.(X+1/Y...