Question

已知ab/a+b=1/3，bc/b+c=1/4，ac/a+c=1/5，求abc/ab+ac+bc的值

Answer 1

ab/(a+b)=1/3(1)
bc/(b+c)=1/4(2)
ac/(a+c)=1/5(3)
abc/(ab+ac+bc)=t
(ab+ac+bc)/abc=1/t
ab/abc+ac/abc+bc/abc=1/t
1/c+1/b+1/a=1/t
(1)两边取倒数：（a+b)/ab=3,a/ab+b/ab=3,1/b+1/a=3(4)
(2)两边取倒数：（b+c)/bc=4,b/bc+c/bc=4,1/c+1/b=4(5)
(3)两边取倒数：（a+c)/ac=5,a/ac+c/ac=5,1/c+1/a=5(6)
(4)+(5)+(6) 1/b+1/a+1/c+1/b+1/c+1/a=3+4+5
2x1/b+1/ax2+1/cx2=12
2(1/b+1/a+1/c)=12
1/a+1/b+1/c=6
6=1/t
t=1/6
abc/(ab+ac+bc)=1/6
答：该表达式是1/6

Answer 2

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