Question

已知x/x2-3x+1=1/5,求x2/x四次方+x2+1的值

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Answer 1

x/x²-3x+1=1/5
x²-3x+1=5x
x²+1=8x
x+1/x=8
平方
x²+2+1/x²=64
x²+1/x²=62

x²/(x^4+x²+1)
=1/(x²+1+1/x²)
=1/63

追问

x²+2+1/x²=64，+2是哪得的？

Answer 2

x/(x^2-3x+1)=1/5
(x^2-3x+1)/x=5
x+1/x-3=5
x+1/x=8
原式=x^2/(x^4+x^2+1)
=1/(x^2+1/x^2+1)
=1/[(x+1/x)^2-1]
=1/(8^2-1)
=1/63