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Question-216105




Question Number 216105 by mr W last updated on 27/Jan/25
Answered by A5T last updated on 27/Jan/25
x^2 =2(((x+1)/2))^2 ⇒x^2 =((x^2 +2x+1)/2)⇒x^2 −2x−1=0  ⇒x=((2+_− (√(4+4)))/2)=1+_− (√2)>0⇒x=1+(√2)
x2=2(x+12)2x2=x2+2x+12x22x1=0x=2+4+42=1+2>0x=1+2
Commented by mr W last updated on 27/Jan/25
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Answered by mehdee7396 last updated on 30/Jan/25
o′(α+1,α)  & A(0,1) & B(α+1,α)  ⇒M(((α+1)/2),((α+1)/2))  C′ : (x−α−1)^2 +y^2 =α^2   M∈C′  ⇒(((α+1)/2)−α−1)^2 +(((α+1)/2))^2 =α^2   ⇒α=(√2)+1 ✓
o(α+1,α)&A(0,1)&B(α+1,α)M(α+12,α+12)C:(xα1)2+y2=α2MC(α+12α1)2+(α+12)2=α2α=2+1
Commented by mehdee7396 last updated on 28/Jan/25

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