x^2-5x+1=0,求代数式(x^2/x-1)-(1+1/x^2-x)的值

来源:百度知道 编辑:UC知道 时间:2024/05/15 16:44:00

x^2-5x+1=0
所以x^2=5x-1
-5x+1=-x^2

x^2/(x-1)-[1+1/(x^2-x)]
=(5x-1)/(x-1)-1-1/(5x-1-x)
=(5x-1)/(x-1)-1-1/(4x-1)
=[(5x-1)(4x-1)-(x-1)(4x-1)-(x-1)]/(x-1)(4x-1)
=(20x^2-9x+1-4x^2+5x-1-x+1)/(4x^2-5x+1)
=(16x^2-5x+1)/(4x^2-5x+1)
=(16x^2-x^2)/(4x^2-x^2)
=15x^2/3x^2
=5

答案是 5
解答:由x^2-5x+1=0,得x+1/x=5
由(x^2/x-1)-(1+1/x^2-x)得x^3/x(x-1)-[(x^2-x+1)/x^2-x]=[x^3-(x^2-x+1)]/x^2-x=(x^2+1)/x=x+1/x=5

答案一定正确!!!