设a=(根号5-1)/2 则(a^5 + a^4 - 2a^3 - a^2 -a +2)/(a^3 - a ) 会等于?
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设a=(根号5-1)/2 则(a^5 + a^4 - 2a^3 - a^2 -a +2)/(a^3 - a ) 会等于?
设a=(根号5-1)/2 则(a^5 + a^4 - 2a^3 - a^2 -a +2)/(a^3 - a ) 会等于?
设a=(根号5-1)/2 则(a^5 + a^4 - 2a^3 - a^2 -a +2)/(a^3 - a ) 会等于?
a=(√5-1)/2,则a+1=(√5+1)/2,所以 a(a+1)=1
(a^5 + a^4 - 2a^3 - a^2 -a +2)/(a^3 - a )
=〔a^3(a+2)(a-1)-(a+2)(a-1)]/[(a(a+1)(a-1)]
=[(a+2)(a-1)(a^2+a+1)]/[a(a+1)]
=(√5+3)/2*(√5-3)/2*[a(a+1)+1]
=-1*2
=-2
(√5-1)/2
(a^5 + a^4 - 2a^3 - a^2 -a +2)/(a^3 - a )
=a³(a²+a-2)-(a²+a-2) /a(a²-1)
=〔a³(a+2)(a-1)-(a+2)(a-1)] /[(a(a+1)(a-1)]
=[(a+2)(a-1)(a²+a+1)] /[a(a+1)]
∵a=(...
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(a^5 + a^4 - 2a^3 - a^2 -a +2)/(a^3 - a )
=a³(a²+a-2)-(a²+a-2) /a(a²-1)
=〔a³(a+2)(a-1)-(a+2)(a-1)] /[(a(a+1)(a-1)]
=[(a+2)(a-1)(a²+a+1)] /[a(a+1)]
∵a=(√5-1)/2 ∴a+1=(√5+1)/2 ∴a(a+1)=1
={(√5+3)/2×(√5-3)/2×[a(a+1)+1]} /1
=[(5-9)/2×2]×[a(a+1)+1]
=-1×2
=-2
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