Question

已知x/x2-3x+1=1/5,求x2;/x4-9x2+1

Answer 1

x/x2-3x+1=1/5,
5x=x²-3x+1
x²-8x+1=0
全部都除以x得到
x-8+1/x=0
x+1/x=8 两边平方
x²+2+1/x²=64
x²+1/x²=62

x²/(x^4-9x²+1）
=1/(x^2-9+1/x^2)
=1/(62-9)
=1/53

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Answer 2

x/(x^2-3x+1)=1/5

x^2-3x+1=5x

x^2+1=8x (等式两边平方)

(x^2+1)^2=64x^2

x^4+2x^2+1=64x^2
x^4+2x^2-11x^2+1=64x^2-11x^2
x^4-9x^2+1=53x^2

x^2/(x^4-9x^2+1)=1/53

Answer 3

x/x2-3x+1=1/5

1/(x-3+1/x)=1/5
x-3+1/x=5

x+1/x=8
x2/(x4-9x2+1)

=1/(x^2-9+1/x^2)
=1/[(x+1/x)^2-2-9]
=1/(64-11)
=1/53