Question Number 104115 by bobhans last updated on 19/Jul/20

Answered by nimnim last updated on 19/Jul/20

((xy)/(x+y))=a, ((xz)/(x+z))=b, ((yz)/(y+z))=c  (1/x)+(1/y)=(1/a),  (1/x)+(1/z)=(1/b),  (1/y)+(1/z)=(1/c)  2((1/x)+(1/y)+(1/z))=(1/a)+(1/b)+(1/c)=((bc+ac+ab)/(abc))  (1/x)+(1/c)=((bc+ac+ab)/(2abc))  (1/x)=((bc+ac+ab)/(2abc))−(1/c)  (1/x)=((bc+ac+ab−2ab)/(2abc))=((bc+ac−ab)/(2abc))  x=((2abc)/(ac+bc−ab))

Commented bybemath last updated on 19/Jul/20

colll

Commented bybobhans last updated on 19/Jul/20

nice !