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

Answered by nimnim last updated on 19/Jul/20

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

Commented by bemath last updated on 19/Jul/20

$$colll$$

Commented by bobhans last updated on 19/Jul/20

$$nice!$$

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