Question

课堂上，李老师给大家出了这样一道题：当x=3，5-2倍根号2，7+根号3时，求代数式x的平方-2x+1/x的平方-1除

Answer 1

当x=3 x的平方-2x+1/x的平方-1=(x-1 )(x-1 )/(x+1 )(x-1 )=(x-1 )/(x+1 )=3-1/(3+1)=1/2
x=5-2倍根号2
x的平方-2x+1/x的平方-1=(x-1 )(x-1 )/(x+1 )(x-1 )=(x-1 )/(x+1 )=4-2根号2/6-2根号2
=2-根号2/3-根号2 =(4-根号2)/7
当x=7+根号3时 x的平方-2x+1/x的平方-1=(x-1 )(x-1 )/(x+1 )(x-1 )=(x-1 )/(x+1 )
=(7+根号3-1)/(7+根号3+1)
=(45+2根号3)/61

Answer 2

(x的平方-2x+1)/(x的平方-1)除以(2x-2)/(x+1)
=(x-1)^2/(x-1)(x+1)÷2(x-1)/(x+1)
=(x-1)/(x+1)÷2(x-1)/(x+1)
=1/2
所以无论x取何值，结果都是1/2
其中x≠1或-1