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Question-191395




Question Number 191395 by Mingma last updated on 23/Apr/23
Answered by Rasheed.Sindhi last updated on 23/Apr/23
xy+yz+zx=0;  ((y+z)/x)+((x+z)/y)+((x+y)/z)=?  ((y+z)/x)+((x+z)/y)+((x+y)/z)  =((y+z)/x)+1+((x+z)/y)+1+((x+y)/z)+1−3  =((x+y+z)/x)+((x+y+z)/y)+((x+y+z)/z)−3  =(x+y+z)((1/x)+(1/y)+(1/z))−3  =(x+y+z)(((xy+yz+zx)/(xyz)))−3  =(x+y+z)((0/(xyz)))−3=−3
$${xy}+{yz}+{zx}=\mathrm{0};\:\:\frac{{y}+{z}}{{x}}+\frac{{x}+{z}}{{y}}+\frac{{x}+{y}}{{z}}=? \\ $$$$\frac{{y}+{z}}{{x}}+\frac{{x}+{z}}{{y}}+\frac{{x}+{y}}{{z}} \\ $$$$=\frac{{y}+{z}}{{x}}+\mathrm{1}+\frac{{x}+{z}}{{y}}+\mathrm{1}+\frac{{x}+{y}}{{z}}+\mathrm{1}−\mathrm{3} \\ $$$$=\frac{{x}+{y}+{z}}{{x}}+\frac{{x}+{y}+{z}}{{y}}+\frac{{x}+{y}+{z}}{{z}}−\mathrm{3} \\ $$$$=\left({x}+{y}+{z}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)−\mathrm{3} \\ $$$$=\left({x}+{y}+{z}\right)\left(\frac{{xy}+{yz}+{zx}}{{xyz}}\right)−\mathrm{3} \\ $$$$=\left({x}+{y}+{z}\right)\left(\frac{\mathrm{0}}{{xyz}}\right)−\mathrm{3}=−\mathrm{3} \\ $$

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