Question Number 192811 by Abdullahrussell last updated on 28/May/23
Answered by AST last updated on 28/May/23
$$\Sigma\frac{\mathrm{1}}{{x}+{yz}}=\Sigma\frac{{x}}{{x}^{\mathrm{2}} +{xyz}}=\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{5}}+\frac{{y}}{{y}^{\mathrm{2}} +\mathrm{5}}+\frac{{z}}{{z}^{\mathrm{2}} +\mathrm{5}} \\ $$$$=\frac{\Sigma\left({xy}^{\mathrm{2}} +\mathrm{5}{x}\right)\left({z}^{\mathrm{2}} +\mathrm{5}\right)=\Sigma\left({xy}^{\mathrm{2}} {z}^{\mathrm{2}} +\mathrm{5}{x}\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+\mathrm{25}{x}\right)}{\mathrm{5}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{x}^{\mathrm{2}} {z}^{\mathrm{2}} \right)+\mathrm{25}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+\mathrm{150}} \\ $$$$\Sigma{x}^{\mathrm{2}} =\left(\Sigma{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\Sigma{xy}\right)=−\mathrm{5};\Sigma{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\left(\Sigma{xy}\right)^{\mathrm{2}} −\mathrm{2}{xyz}\left(\Sigma{x}\right)=−\mathrm{1} \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{{x}+{yz}}=\frac{{xyz}\left(\Sigma{xy}\right)+\mathrm{25}\left(\Sigma{x}\right)+\left(\Sigma{x}\right)\left(\Sigma{xy}\right)−\mathrm{3}{xyz}}{\mathrm{20}} \\ $$$$=\frac{\mathrm{5}\left(\mathrm{3}\right)+\mathrm{25}\left(\mathrm{1}\right)+\mathrm{1}\left(\mathrm{3}\right)−\mathrm{15}=\mathrm{28}}{\mathrm{20}}=\frac{\mathrm{7}}{\mathrm{5}} \\ $$