a{1/b+1/c}+b{1/c+1/a}+c{1/a+1/b}+3
=a/b+a/c+b/c+b/a+c/a+c/b+3
=[a(b+c)+b(a+c)+c(a+b)]/abc+3
=(2ab+2ac+2bc)/abc+3.①
∵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+3
=3
∴原式=3
暴力解法
帶a=-1;b=-1;c=2進(jìn)去算