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Question-164024

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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
x2−y2=−(x−y)(x−y)(x+y)+(x−y)=0(x−y)(x+y+1)=0x+y=−1[x≠y]x2+y2=x+y+6x2+y2−1=−1−1+6x2+y2−1=2

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