Question

已知1/a+1/b+1/c=1/(a+b+c),求(a+b)(a+c)(b+c)=0

Answer 1

1/a+1/b+1/c－1/(a+b+c)＝0

(a＋b)/(ab)+(a＋b)/[c(a+b+c)]＝0

(a＋b){1/(ab)+1/[c(a+b+c)]}＝0

(a＋b){[c(a＋b＋c)＋ab]/[abc(a+b+c)]}＝0

(a＋b)(ac＋bc＋c^2＋ab)＝0

(a＋b)[c^2＋(a＋b)c＋ab]＝0

(a＋b)(b＋c)(c＋a)＝0