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soit-le-systeme-d-equatiins-x-y-z-7-x-2-y-2-z-2-9-xyz-5-1-x-1-y-1-z-




Question Number 211368 by a.lgnaoui last updated on 07/Sep/24
soit le systeme d equatiins    x+y+z    =7   x^2 +y^2 +z^2 =9    xyz            =5    (1/x)+(1/y)+(1/z)?
$$\mathrm{soit}\:\mathrm{le}\:\mathrm{systeme}\:\mathrm{d}\:\mathrm{equatiins} \\ $$$$\:\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}\:\:\:\:=\mathrm{7} \\ $$$$\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\boldsymbol{\mathrm{z}}^{\mathrm{2}} =\mathrm{9} \\ $$$$\:\:\boldsymbol{\mathrm{xyz}}\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{5} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}}? \\ $$
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)=((7^2 −9)/2)=20  ∴((1/x)+(1/y)+(1/z))  =((xy+gz+zx)/(xyz))  =((20)/5)=4
$$\:\mathrm{2}\left({xy}+{yz}+{zx}\right)=\left({x}+{y}+{z}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right) \\ $$$$\:\left({xy}+{yz}+{zx}\right)=\frac{\mathrm{7}^{\mathrm{2}} −\mathrm{9}}{\mathrm{2}}=\mathrm{20} \\ $$$$\therefore\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right) \\ $$$$=\frac{{xy}+{gz}+{zx}}{{xyz}} \\ $$$$=\frac{\mathrm{20}}{\mathrm{5}}=\mathrm{4} \\ $$

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