Question

已知：ab/(a+b）=1/3，bc/(b+c)=1/4，ac/(a+c)=1/5，求abc/(a+b+c)。

Answer 1

ab/a+b=1/3,bc/b+c=1/4,ac/a+c=1/5可以取倒数得出a+b/ab=3,b+c/bc=4,a+c/ac=5分解有
1/a+1/b=3,1/b+1/c=4,1/a+1/c=5,解得
a=1/2,b=1,c=1/3由此可得
1/a+1/b+1/c=6得
abc/ab+bc+ca=1/(1/a+1/b+1/c)=1/6

Answer 2

ab/a+b=1/3,bc/b+c=1/4,ac/a+c=1/5
取倒数得出a+b/ab=3,b+c/bc=4,a+c/ac=5
1/a+1/b=3,1/b+1/c=4,1/a+1/c=5,解得
a=1/2,b=1,c=1/3由此可得
abc/(a+b+c)=(1/6)/(1/2+1+1/3)=1/11

Answer 3

ab+bc+ac/abc=1/a+1/b+1/c =6
所以abc/(ab+bc+ac)=1/6

Answer 4

1 (a+b)/ab=3
1/b+1/a=3
2 (b+c)/bc=4
1/c+1/b=4
3 (a+c)/ac=5
1/c+1/a=5
1/b+1/a+1/c+1/b+1/c+1/a=3+4+5
2(1/a+1/b+1/c)=12
1/a+1/b+1/c=6
(a+b+c)/abc=6
abc/(a+b+c)=1/6