已知x平方-x-1=0,则x四次方+2x+1/x五次方=？
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x²-x-1=0
x²=x+1
(x⁴+2x+1)/x^5
=[(x²)²+2x+1]/[x(x²)²]
=[(x+1)²+2x+1]/[x(x+1)²]
=(x²+4x+2)/[x(x²+2x+1)]
=(x+1+4x+2)/[x(x+1+2x+1)]
=(5x+3)/[x(3x+2)]
=(5x+3)/(3x²+2x)
=(5x+3)/[3(x+1)+2x]
=(5x+3)/(5x+3)
=1
提示：就是反复用x+1代替x².,6,这题关键是逐步找到（x^2-x-1)
还有x^2-x=1,2,
x²=x+1
(x⁴+2x+1)/x^5
=[(x²)²+2x+1]/[x(x²)²]
=[(x+1)²+2x+1]/[x(x+1)²]
=(x²+4x+2)/[x(x²+2x+1)]
=(x+1+4x+2)/[x(x+1+2x+1)]
=(5x+3)/[x(3x+2)]
=(5x+3)/(3x²+2x)
=(5x+3)/[3(x+1)+2x]
=(5x+3)/(5x+3)
=1
提示：就是反复用x+1代替x².,6,这题关键是逐步找到（x^2-x-1)
还有x^2-x=1,2,
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