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