Question

ab/a+b=3 ac/a+c=4 bc/b+c=5,求abc/a+b+c

Answer 1

已知ab/(a+b)=3,ac/(a+c)=4,bc/(b+c)=5,求abc/(a+b+c)
(a+b+c)/abc=1/bc+1/ac+1/ab.(1)
由ab/(a+b)=3,得(a+b)/ab=1/3,即有1/a+1/b=1/3.(2)
由ac/(a+c)=4,得(a+c)/ac=1/4,即有1/a+1/c=1/4.(3)
由bc/(b+c)=5,得(b+c)/bc=1/5,即有1/b+1/c=1/5.(4)
(2)-(3)得1/b-1/c=1/3-1/4=1/12.(5)
(4)+(5)得2/b=1/5+1/12=17/60,故1/b=17/120.(6)
(2)-(4)得1/a-1/c=1/3-1/5=2/15.(7)
(3)+(7)得2/a=1/4+2/15=23/60,故1/a=23/120.(8)
将(8)代入(3)式得1/c=1/4-1/a=1/4-23/120=7/120.(9)
(6)×(9)得1/bc=(17/120)×(7/120)=119/14400.(10)
(8)×(9)得1/ac=(23/120)×(7/120)=161/14400.(11)
(8)×(6)得1/ab=(23/120)×(17/120)=391/14400.(12)
将(10)(11)(12)代入(1)式得(a+b+c)/abc=(119+161+391)/14400=671/14400
∴abc/(a+b+c)=14400/671