一道分式数学题、已知x^2+x-1=0、求x-[1-2/(1-x)]/（x+1)-[x(x^2-1)]/x^2-2x+1的值

一道分式数学题、已知x^2+x-1=0、求x-[1-2/(1-x)]/（x+1)-[x(x^2-1)]/x^2-2x+1的值
一道分式数学题、
已知x^2+x-1=0、求x-[1-2/(1-x)]/（x+1)-[x(x^2-1)]/x^2-2x+1的值

一道分式数学题、已知x^2+x-1=0、求x-[1-2/(1-x)]/（x+1)-[x(x^2-1)]/x^2-2x+1的值
x{[1-2/(1-x)]/(x+1)-(x^2-1)/(x^2-2x+1)}
=x{[-(x+1)/(1-x)]/(x+1)-(x-1)(x+1)/(x-1)^2}
=x{-1/(1-x)-(x+1)/(x-1)}
=-x^2/(x-1),
因为x^2+x-1=0,
所以x^2=-(x-1),
所以原式=1.

x-[1-2/(1-x)]/（x+1)-[x(x^2-1)]/x^2-2x+1
=x-[(1-x-2)/(1-x)]/（x+1)-[x(x+1)(x-1)]/(x-1)^2
=x-(x+1))/(x+1)(x-1)-x(x+1)/(x-1)
=x-1/(x-1)-x(x+1)/(x-1)
=(x^2-1-1-x^2-x)/(x-1)
=-(x+2)/(x-1)
你写错了吧
前面真是x- ？

请问你那个2是X的2A平方吗？确定你打出来的式子没错？