已知χ满足χ²-3χ+1=0,求χ²/χ⁴-χ²+1的值

已知χ满足χ²-3χ+1=0,求χ²/χ⁴-χ²+1的值

χ²-3χ+1=0,
得χ²=3χ-1
逐步代入降次
χ⁴=（χ²）²==（3χ-1）²=9χ²-6χ+1
χ⁴-χ²+1=（9χ²-6χ+1）-χ²+1=（8χ²-6χ+2）=18x-6
χ²/χ⁴-χ²+1=(3χ-1)/(18x-6)=1/6

x^2-3x+1=0
x^2=3x-1

原式=x^2/[(3x-1)^2-(3x-1)+1]
=x^2/[3(3x^2-3x+1)]
=x^2/[3(x^2-3x+1+2x^2)]
=x^2/(6x^2)
=1/6

【典型解法】
χ²-3χ+1=0
方程除以x
x+1/x=3
两边平方
x^2+2+1/x^2=9
x^2+1/x^2=7
则
χ²/χ⁴-χ²+1
=1/（x^2+1/x^2-1)
=1/6