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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))

$$\frac{{xy}}{{x}+{y}}={a},\:\frac{{xz}}{{x}+{z}}={b},\:\frac{{yz}}{{y}+{z}}={c} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{{a}},\:\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{b}},\:\:\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{c}} \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{{bc}+{ac}+{ab}}{{abc}} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{c}}=\frac{{bc}+{ac}+{ab}}{\mathrm{2}{abc}} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{{bc}+{ac}+{ab}}{\mathrm{2}{abc}}−\frac{\mathrm{1}}{{c}} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{{bc}+{ac}+{ab}−\mathrm{2}{ab}}{\mathrm{2}{abc}}=\frac{{bc}+{ac}−{ab}}{\mathrm{2}{abc}} \\ $$$${x}=\frac{\mathrm{2}{abc}}{{ac}+{bc}−{ab}} \\ $$

Commented by bemath last updated on 19/Jul/20

colll

$${colll} \\ $$

Commented by bobhans last updated on 19/Jul/20

nice !

$${nice}\:! \\ $$

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