Question

已知x^2-5x+1=0,求x^2/x^4-3x^2+1的值

已知x^2-5x+1=0,求x^2/x^4-3x^2+1的值

Answer 1

解析：
已知x²-5x+1=0，那么：x²+1=5x且可知x≠0
上式两边同除以x可得：x+ 1/x=5
两边平方得：(x+ 1/x)²=25
即：x²+2+ 1/x²=25
x²+ 1/x²=23
所以：x²/(x^4 -3x²+1） （分子分母同除以x²得到下式）
=1/(x² - 3 + 1/x²)
=1/(23-3)
=1/20

Answer 2

x^2-5x+1=0
x+1/x=5
x^2+1/x^2=23

x^4-3x^2+1/x^2
=x^2-3+1/x^2
=23-3
=20

x^2/(x^4-3x^2+1)=1/20

Answer 3

俊狼猎英团队为您解答：

降次法：
由：x^2-5x+1=0,得X^2=5X-1
x^2/x^4-3x^2+1
=X^2/［(5X-1)^2-3(5X-1)+1］
=X^2/［25X^2-25X+5］
=(5X-1)/［5(25X-5-5X+1)］
=(5X-1)/［5(20X-4)］
=1/20

Answer 4

x^2=5x-1
x^4-3x^2+1
=（5x-1）^2-3x^2+1
=22x^2-10x+2
=22(5x-1)-10x+2
=100x-20=20(5x-1)
所以原式子=(5x-1)/[20(5x-1)]=1/20

Answer 5

2/7+3根号5