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Question Number 192811 by Abdullahrussell last updated on 28/May/23

Answered by deleteduser1 last updated on 28/May/23

$$\mathrm{\Sigma }\frac{1}{x+yz}=\mathrm{\Sigma }\frac{x}{x^2+xyz}=\frac{x}{x^2+5}+\frac{y}{y^2+5}+\frac{z}{z^2+5}$$$$=\frac{\mathrm{\Sigma }({xy}^2+5x)(z^2+5)=\mathrm{\Sigma }({xy}^2{z}^2+5x(y^2+z^2)+25x)}{5(x^2{y}^2+y^2{z}^2+x^2{z}^2)+25(x^2+y^2+z^2)+150}$$$$\mathrm{\Sigma }{x}^2={\left(\mathrm{\Sigma }x\right)}^2-2\left(\mathrm{\Sigma }xy\right)=-5;\mathrm{\Sigma }{x}^2{y}^2={\left(\mathrm{\Sigma }xy\right)}^2-2xyz\left(\mathrm{\Sigma }x\right)=-1$$$$\Rightarrow \mathrm{\Sigma }\frac{1}{x+yz}=\frac{xyz\left(\mathrm{\Sigma }xy\right)+25\left(\mathrm{\Sigma }x\right)+\left(\mathrm{\Sigma }x\right)\left(\mathrm{\Sigma }xy\right)-3xyz}{20}$$$$=\frac{5\left(3\right)+25\left(1\right)+1\left(3\right)-15=28}{20}=\frac{7}{5}$$

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