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x-1-x-1-x-1907-1-x-1907-




Question Number 102771 by bramlex last updated on 11/Jul/20
x+(1/x) = −1 ⇒x^(1907) +(1/x^(1907) ) ?
x+1x=1x1907+1x1907?
Answered by ajfour last updated on 11/Jul/20
x^2 +x+1=0  ⇒  x=ω, ω^2     ω^(1907) +(1/ω^(1907) )= (1/ω)+ω = ω^2 +ω=−1.
x2+x+1=0x=ω,ω2ω1907+1ω1907=1ω+ω=ω2+ω=1.
Answered by floor(10²Eta[1]) last updated on 11/Jul/20
x^2 =−x−1  x^4 =x^2 +2x+1=(−x−1)+2x+1=x  ⇒x^4 =x⇒x^(16) =x^4 =x  x^(160) =x^(10) =x^4 .x^4 .x^2 =x.x.x^2 =x^4 =x  ⇒x^(160) =x⇒x^(1600) =x^(10) =x  ★x^(307) =(x^(10) )^(30) .x^7 =x^(30) .x^4 .x^3 =(x^(10) )^3 .x.x^3   =x^3 .x^4 =x^3 .x=x  ⇒x^(1907) =x^(1600) .x^(307) =x.x=x^2   x^(1907) +(1/x^(1907) )=x^2 +(1/x^2 )=((x^4 +1)/x^2 )=((x+1)/(−x−1))=−1
x2=x1x4=x2+2x+1=(x1)+2x+1=xx4=xx16=x4=xx160=x10=x4.x4.x2=x.x.x2=x4=xx160=xx1600=x10=xx307=(x10)30.x7=x30.x4.x3=(x10)3.x.x3=x3.x4=x3.x=xx1907=x1600.x307=x.x=x2x1907+1x1907=x2+1x2=x4+1x2=x+1x1=1
Commented by Rasheed.Sindhi last updated on 11/Jul/20
Cool!
Cool!
Commented by bramlex last updated on 11/Jul/20
jooss...
jooss

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