Question

已知；a+b+c=0,且abc不等于0，计算a{1/b+1/c}+b{1/c+1/a}+c{1/a+1/b}+3的值,帮帮忙

Answer 1

解：abc不等于0,且a+b+c=0
所以
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+b/c)+(b/a+c/a)+3
=(a+c)/b+(a+b)/c+(b+c)/a+3
=-b/b-c/c-a/a+3
=-1-1-1+3
=0

Answer 2

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

带a=-1;b=-1;c=2进去算

Answer 3

0