Question

已知a＋b＋c＝0,1／a＋1／b＋1／c＝﹣4,那么1／a²＋1／b²＋1／c²的值为

Answer 1

可化为1/(a^2)+1/(b^2)+1/(c^2)
=(a^2b^2+b^2c^2+c^2a^2)/(a^2b^2c^2)
因为（1/a+1/b+1/c）^2=(a^2b^2+2ab^2c+b^2c^2+2bc^2a+c^2a^2+2ca^b)/(a^2b^2c^2)
比上面多了（2ab^2c+2bc^2a+2ca^b）/(a^2b^2c^2)=2(a+b+c)/(abc)〉〉
所以原式
=（1/a+1/b+1/c）^2-2(a+b+c)/(abc)
=16-0
=16

Answer 2

(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2/ab+2/ac+2/bc
(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2(a+b+c)/abc
1/a^2+1/b^2+1/c^2=(1/a+1/b+1/c)^2-2(a+b+c)/abc
=(-4)^2-0
=16

Answer 3

(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2/ab+2/ac+2/bc
(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2(a+b+c)/abc
1/a^2+1/b^2+1/c^2=(1/a+1/b+1/c)^2-2(a+b+c)/abc
=(-4)^2-0
=16