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Question-140688




Question Number 140688 by mohammad17 last updated on 11/May/21
Commented by mohammad17 last updated on 11/May/21
help me
$${help}\:{me} \\ $$
Answered by TheSupreme last updated on 11/May/21
(1/x)+(1/y)+(1/z)=((xy+xz+yz)/(xyz))=(1/2)((2(xy+xz+yz)+x^2 +y^2 +z^2 −x^2 −y^2 −z^2 =)/(xyz))  =(1/2)(((x+y+z)^2 −(x^2 +y^2 +z^2 ))/(xyz))=(1/2)((7^2 −9)/5)=4
$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{{xy}+{xz}+{yz}}{{xyz}}=\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{2}\left({xy}+{xz}+{yz}\right)+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} −{z}^{\mathrm{2}} =}{{xyz}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{{xyz}}=\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{7}^{\mathrm{2}} −\mathrm{9}}{\mathrm{5}}=\mathrm{4} \\ $$
Answered by som(math1967) last updated on 11/May/21
 (x+y+z)^2 =7^2   x^2 +y^2 +z^2 +2(xy+yz+zx)=49  (xy+yz+zx)=((49−9)/2) =20 ★  (1/x)+(1/y)+(1/z)  =((yz+zx+xy)/(xyz))=((20)/5)=4 ans  ∵x^2 +y^2 +z^2 =9
$$\:\left({x}+{y}+{z}\right)^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right)=\mathrm{49} \\ $$$$\left({xy}+{yz}+{zx}\right)=\frac{\mathrm{49}−\mathrm{9}}{\mathrm{2}}\:=\mathrm{20}\:\bigstar \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}} \\ $$$$=\frac{{yz}+{zx}+{xy}}{{xyz}}=\frac{\mathrm{20}}{\mathrm{5}}=\mathrm{4}\:\boldsymbol{{ans}} \\ $$$$\because\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} =\mathrm{9} \\ $$$$ \\ $$

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