Question

已知a是方程x^2+x-1=0的一个根,则2/(a^2-1)-1/(a^2-a)的值为（ ） 解题步骤要有

Answer 1

即a²+a-1=0
则a²-1=a
a²-a=-2a+1
2a²-a=-3a+1
所以原式=2/a-1/(-2a+1)
=2/a+1/(2a-1)
=(4a-1+a)/(2a²-a)
=(3a-1)/(-3a+1)
=-1

Answer 2

x²+x-1=0
x²+x-1/2-1/2=0
(x-1/2)²=1/2
x-1/2=±根号1/2
x=1/2-根号1/2或x=1/2+根号1/2
2/(a²-1)-1/(a²-a)
=2a/(a+1)(a-1）—(a+1)/a(a-1)(a+1)
=2a/(a+1)(a-1)—(a+1)/a(a-1)(a+1)
=2a-a-1/a(a-1)(a+1)
=a-1/a(a-1)(a+1)
=1/a(a+1)
当a=1/2-根号1/2时
原式=1/a(a+1)
=1/[(1/2-根号1/2)²+1/2-根号1/2]
=1/(5/4-根号2)
=4/(5-4根号2)
当a=1/2+根号1/2时
原式=1/a(a+1)
=1/[(1/2+根号1/2)²+1/2+根号1/2]
=1/(1+ 根号2)