Question Number 164024 by mathlove last updated on 13/Jan/22
Answered by som(math1967) last updated on 13/Jan/22
![x^2 −y^2 =−(x−y) (x−y)(x+y)+(x−y)=0 (x−y)(x+y+1)=0 x+y=−1 [x≠y] x^2 +y^2 =x+y+6 x^2 +y^2 −1=−1−1+6 (√(x^2 +y^2 −1))=2](https://www.tinkutara.com/question/Q164029.png)
$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =−\left({x}−{y}\right) \\ $$$$\left({x}−{y}\right)\left({x}+{y}\right)+\left({x}−{y}\right)=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}+{y}=−\mathrm{1}\:\left[{x}\neq{y}\right] \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={x}+{y}+\mathrm{6} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}=−\mathrm{1}−\mathrm{1}+\mathrm{6} \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2} \\ $$$$ \\ $$