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Question Number 211368 by a.lgnaoui last updated on 07/Sep/24

$$\mathrm{soit}\mathrm{le}\mathrm{systeme}\mathrm{d}\mathrm{equatiins}$$$$\mathbf{x}+\mathbf{y}+\mathbf{z}=7$$$${\mathbf{x}}^2+{\mathbf{y}}^2+{\mathbf{z}}^2=9$$$$\mathbf{xyz}=5$$$$$$$$\frac{1}{\mathbf{x}}+\frac{1}{\mathbf{y}}+\frac{1}{\mathbf{z}}?$$

Answered by som(math1967) last updated on 07/Sep/24

$$2(xy+yz+zx)={(x+y+z)}^2-(x^2+y^2+z^2)$$$$(xy+yz+zx)=\frac{7^2-9}{2}=20$$$$\therefore (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$$$$=\frac{xy+gz+zx}{xyz}$$$$=\frac{20}{5}=4$$

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