设X

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设X

设X
设X

设X
x-1/x=√5
(x^10+x^6+x^4+1)/(x^10+x^8+x^2+1)=?
∵(x+1/x)^2=(x-1/x)^2+4=9
∴x+1/x=-3
x^2+1/x^2=(x+1/x)^2-2=7
x^3+1/x^3=(x+1/x)^3-3(x+1/x)=-27-3(-3)=-18
x^4+1/x^4=(x^2+1/x^2)^2-2=7^2-2=47
x^5+1/x^5=(x^3+1/x^3)(x^2+1/x^2)-(x+1/x)=-18×7-(-3)=-123
将原式分子分母同除以x^5
原式=(x^5+x+1/x+1/x^5)/(x^5+x^3+1/x^3+1/x^5)=(-123-3)/(-123-18)=126/141

x-1/x=√5两边平方
得x^2+1/x^2-2=5
即x^2+1/x^2=7
即x^4+1=7x^2
(x^10+x^6+x^4+1)/(x^10+x^8+x^2+1)
=[x^6(x^4+1)+(x^4+1)]/[x8(x^2+1)+(x^2+1)]
=[(x^6+1)(x^4+1)]/[(x^8+1)(x^2+1)]
=[(x^2+1...

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x-1/x=√5两边平方
得x^2+1/x^2-2=5
即x^2+1/x^2=7
即x^4+1=7x^2
(x^10+x^6+x^4+1)/(x^10+x^8+x^2+1)
=[x^6(x^4+1)+(x^4+1)]/[x8(x^2+1)+(x^2+1)]
=[(x^6+1)(x^4+1)]/[(x^8+1)(x^2+1)]
=[(x^2+1)(x^4+1-x^2)(x^4+1)]/[(x^8+1)(x^2+1)]
=6x^2*7x^2/(x^8+1)
=42/(x^4+1/x^4)
=42/[(x^2+1/x^2)^2-2]
=42/(7^2-2)
=42/47

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